$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdot cdotfrac{99}{100})<frac{1}{10}$.
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Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
inequality
|
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up vote
6
down vote
favorite
Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
inequality
3
Related
– Kemono Chen
yesterday
2
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
yesterday
2
@KemonoChen I think the correct inequality should be $$frac{2}{3sqrt{2n}}<frac12cdotfrac34cdotldotscdotfrac{2n-1}{2n}<frac{1}{sqrt{2n}},.$$ Otherwise, the left-hand side inequality is not even true for $n=1$ (and it is never true), while the right-hand side inequality is very weak.
– Batominovski
yesterday
3
$$frac{1}{4^n}binom{2n}{n}approx frac{1}{sqrt{pileft(n+frac{1}{4}right)}}.$$
– Jack D'Aurizio
yesterday
2
I found that $$frac{1}{sqrt{pi left(n+frac12right)}}<prod_{k=1}^nleft(frac{2k-1}{2k}right)<frac{1}{sqrt{pi n}}$$ with @JackD'Aurizio's estimate being extremely close. For $n=50$, the required product is between $0.07939248$ and $0.07979788$, with Jack's estimate being $0.07958972$. The actual value is $0.07958923$.
– Zvi
yesterday
|
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
inequality
Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
inequality
inequality
edited yesterday
dmtri
1,2261521
1,2261521
asked yesterday
Lovro Sindičić
249216
249216
3
Related
– Kemono Chen
yesterday
2
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
yesterday
2
@KemonoChen I think the correct inequality should be $$frac{2}{3sqrt{2n}}<frac12cdotfrac34cdotldotscdotfrac{2n-1}{2n}<frac{1}{sqrt{2n}},.$$ Otherwise, the left-hand side inequality is not even true for $n=1$ (and it is never true), while the right-hand side inequality is very weak.
– Batominovski
yesterday
3
$$frac{1}{4^n}binom{2n}{n}approx frac{1}{sqrt{pileft(n+frac{1}{4}right)}}.$$
– Jack D'Aurizio
yesterday
2
I found that $$frac{1}{sqrt{pi left(n+frac12right)}}<prod_{k=1}^nleft(frac{2k-1}{2k}right)<frac{1}{sqrt{pi n}}$$ with @JackD'Aurizio's estimate being extremely close. For $n=50$, the required product is between $0.07939248$ and $0.07979788$, with Jack's estimate being $0.07958972$. The actual value is $0.07958923$.
– Zvi
yesterday
|
show 1 more comment
3
Related
– Kemono Chen
yesterday
2
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
yesterday
2
@KemonoChen I think the correct inequality should be $$frac{2}{3sqrt{2n}}<frac12cdotfrac34cdotldotscdotfrac{2n-1}{2n}<frac{1}{sqrt{2n}},.$$ Otherwise, the left-hand side inequality is not even true for $n=1$ (and it is never true), while the right-hand side inequality is very weak.
– Batominovski
yesterday
3
$$frac{1}{4^n}binom{2n}{n}approx frac{1}{sqrt{pileft(n+frac{1}{4}right)}}.$$
– Jack D'Aurizio
yesterday
2
I found that $$frac{1}{sqrt{pi left(n+frac12right)}}<prod_{k=1}^nleft(frac{2k-1}{2k}right)<frac{1}{sqrt{pi n}}$$ with @JackD'Aurizio's estimate being extremely close. For $n=50$, the required product is between $0.07939248$ and $0.07979788$, with Jack's estimate being $0.07958972$. The actual value is $0.07958923$.
– Zvi
yesterday
3
3
Related
– Kemono Chen
yesterday
Related
– Kemono Chen
yesterday
2
2
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
yesterday
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
yesterday
2
2
@KemonoChen I think the correct inequality should be $$frac{2}{3sqrt{2n}}<frac12cdotfrac34cdotldotscdotfrac{2n-1}{2n}<frac{1}{sqrt{2n}},.$$ Otherwise, the left-hand side inequality is not even true for $n=1$ (and it is never true), while the right-hand side inequality is very weak.
– Batominovski
yesterday
@KemonoChen I think the correct inequality should be $$frac{2}{3sqrt{2n}}<frac12cdotfrac34cdotldotscdotfrac{2n-1}{2n}<frac{1}{sqrt{2n}},.$$ Otherwise, the left-hand side inequality is not even true for $n=1$ (and it is never true), while the right-hand side inequality is very weak.
– Batominovski
yesterday
3
3
$$frac{1}{4^n}binom{2n}{n}approx frac{1}{sqrt{pileft(n+frac{1}{4}right)}}.$$
– Jack D'Aurizio
yesterday
$$frac{1}{4^n}binom{2n}{n}approx frac{1}{sqrt{pileft(n+frac{1}{4}right)}}.$$
– Jack D'Aurizio
yesterday
2
2
I found that $$frac{1}{sqrt{pi left(n+frac12right)}}<prod_{k=1}^nleft(frac{2k-1}{2k}right)<frac{1}{sqrt{pi n}}$$ with @JackD'Aurizio's estimate being extremely close. For $n=50$, the required product is between $0.07939248$ and $0.07979788$, with Jack's estimate being $0.07958972$. The actual value is $0.07958923$.
– Zvi
yesterday
I found that $$frac{1}{sqrt{pi left(n+frac12right)}}<prod_{k=1}^nleft(frac{2k-1}{2k}right)<frac{1}{sqrt{pi n}}$$ with @JackD'Aurizio's estimate being extremely close. For $n=50$, the required product is between $0.07939248$ and $0.07979788$, with Jack's estimate being $0.07958972$. The actual value is $0.07958923$.
– Zvi
yesterday
|
show 1 more comment
5 Answers
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Let $P_n:=prodlimits_{k=1}^n,dfrac{2k-1}{2k}$ for each integer $ngeq 1$. We shall prove that $$frac{2}{3sqrt{2n}}<P_n<frac{1}{sqrt{2n}}tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $dfrac{1}{sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$). The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2leq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k}{2k+1}right)=frac{1}{4},prod_{k=3}^{2n},frac{k}{k+1}=frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2geq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k-2}{2k-1}right)=frac{1}{4},prod_{k=2}^{2n-1},frac{k}{k+1}=frac{2}{4(2n)},.$$
This shows that
$$frac{1}{2,sqrt{n}}leq P_nleq frac{sqrt{3}}{2,sqrt{2n+1}},.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$. Note that this implies (*). In particular, for $n=50$, we have
$$frac{1}{15}<0.0707<frac{1}{2cdotsqrt{50}}<P_{50}<frac{sqrt{3}}{2cdot sqrt{101}}<0.0862<frac{1}{10},.$$
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2
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Let $S_n=prod_{k=1}^nfrac{2k-1}{2k}$. From Wallis' product, we have
$$prod_{k=1}^inftyleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)=frac{pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=1}^nleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)<frac{pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)prod_{k=1}^nleft(frac{2k-1}{2k}right)^2=prod_{k=1}^nleft(frac{2k-1}{2k}cdotfrac{2k+1}{2k}right)>frac{2}{pi}.$$
Therefore,
$$S_n>frac{1}{sqrt{pileft(n+frac12right)}}.$$
Similarly, Wallis' product also implies that
$$prod_{k=2}^inftyleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)=frac{4}{pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac{4}{pi}$$
for all $n$. That is,
$$2nS_n^2=2nprod_{k=1}^nleft(frac{2k-1}{2k}right)^2=frac12prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac12left(frac{4}{pi}right).$$
Therefore,
$$S_n<frac{1}{sqrt{pi n}}.$$ Hence,
$$frac{1}{sqrt{pileft(n+frac12right)}}<S_n<frac1{sqrt{pi n}}$$
for every $n$.
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2
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Let $a=frac{1}{2}.frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
and:
$b=frac{2}{3}.frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$frac{1}{2}<frac{2}{3}, frac{3}{4}<frac{4}{5}. . . frac{99}{100}<frac{100}{101}$
⇒ $a^2 < ab=(frac{1}{2}.frac{2}{3}).(frac{3}{4}.frac{4}{5}). . . .(frac{99}{100}.frac{100}{101})=frac{1}{101}$
$a^2<frac{1}{101}$⇒$a<frac{1}{sqrt{101}}<frac{1}{10}$
Also:
$2a=frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
$frac{3}{2}.b=frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
$2a<frac{3}{2}b$⇒$2a^2<frac{3}{2}ab=frac{3}{2}.frac{3}{101}$
Or $a^2<frac{3}{4}ab=frac{3}{4}.frac{3}{101}$
Since $9>4$ then $a^2 >frac{4}{9times 101}$ and therefore:
$a>frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=frac{1}{101}$⇒$a^2>frac{1}{202}$⇒$a>frac{1}{sqrt{202}}>frac{1}{sqrt{225}}=frac{1}{15}$
2
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
3
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
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Just a thought, that may be worth mentioning:
The expression:
$$p=(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$frac{1.3.5.7}{2.4.6.8}=frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=frac{odd_numbers}{even_numbers}=frac{p_o}{p_e}=$$
$$p=frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<frac{(n!)}{((n!)2^{n})^2}$$
$$p<frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
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This is tedious and unsophisticated, but don't knock it, it works! :)
As well as answering the present question, it also answers the linked question, prove $frac{1}{13}<frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}<frac{1}{12}$.
We need a preliminary lemma:
If
$$
k = left(1 - frac{1}{52^2}right)left(1 - frac{1}{56^2}right)cdotsleft(1 - frac{1}{96^2}right),
$$
then
$$
frac{383}{384} < k < frac{1300}{1303}.
$$
Rounding up and down, as appropriate, this is approximately
$$
0.997395 < k < 0.997698,
$$
but of course we avoid using such calculations.
Proof. By the Weierstrass product inequality, we have
$$
1 - s < k < frac{1}{1 + s},
$$
where
$$
s = frac{1}{52^2} + frac{1}{56^2} + cdots + frac{1}{96^2} =
frac{1}{16}left(frac{1}{13^2} + frac{1}{14^2} + cdots + frac{1}{24^2}right).
$$
Telescoping,
begin{align*}
16s & <
frac{1}{12cdot13} + frac{1}{13cdot14} + cdots + frac{1}{23cdot24}
= frac{1}{24}, \
16s & >
frac{1}{13cdot14} + frac{1}{14cdot15} + cdots + frac{1}{24cdot25}
= frac{12}{325},
end{align*}
therefore
$$
1 - frac{1}{384} < k < frac{1}{1 + frac{3}{1300}},
$$
as required. $square$
The number we wish to approximate is
$$
P = frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}
= frac{51cdot53cdot55cdots97cdot99}{2^{25}cdot4cdot8cdot12cdots96cdot100}
= kQ,
$$
where
begin{gather*}
Q = frac{52^2cdot56^2cdots96^2cdot99}{2^{75}cdot25!}
= frac{13cdot14cdots24cdot99}{2^{27}cdot12!cdot25}
= frac{13cdot15cdot17cdot19cdot21cdot23cdot99}{2^{21}cdot6!cdot25} \
= frac{7cdot13cdot17cdot19cdot23cdot99}{2^{25}cdot25}.
end{gather*}
Therefore, using the bounds obtained for $k$ in the lemma,
begin{equation}
tag{$1$}label{ineq:1}
frac{7cdot13cdot17cdot19cdot23cdot33cdot383}{2^{32}cdot25}
< P <
frac{7cdot13^2cdot17cdot19cdot23cdot99}{2^{23}cdot1303}.
end{equation}
Although this is well within the range of easy hand calculation, I lazily used a calculator for most of the work (only doubling $2^{22}cdot1303 = 5{,}465{,}178{,}112$ at the last stage by hand), to get
begin{equation}
tag{$2$}label{ineq:2}
frac{8{,}544{,}456{,}921}{107{,}374{,}182{,}400} < P < frac{870{,}062{,}193}{10{,}930{,}356{,}224}.
end{equation}
Approximately, rounding up and down again,
$$
0.079576 < P < 0.079601.
$$
One way to "simplify" eqref{ineq:1} (although it seems simpler to me to calculate eqref{ineq:2} and be done with it!) is as follows:
For the upper bound, observe that $7cdot13cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2cdot3cdot7cdot31$, so
$$
frac{7cdot13cdot19}{1303} < frac{1728}{2cdot3cdot7cdot31} = frac{288}{217} < frac{288}{216} = frac{4}{3},
$$
(as one can now easily verify with hindsight), therefore
$$
P < frac{13cdot17cdot23cdot99}{2^{21}cdot3} < frac{17cdot300cdot100}{2^{11}cdot3cdot1000} = frac{85}{1024} < frac{1}{12}.
$$
For the lower bound, first simplify $Q$ by noticing that $7cdot17cdot19cdot23 = 52003 > 52000 = 2^5cdot5^3cdot13$, whence
$$
Q > frac{5cdot13^2cdot99}{2^{20}} = frac{5cdot13cdot1287}{2^{20}} > frac{5cdot13cdot1280}{2^{20}} = frac{5^2cdot13}{2^{12}},
$$
and then, instead of using the lower bound from the lemma directly, weaken it to $P > frac{64}{65}Q$, which gives
$$
P > frac{5}{64} > frac{1}{13}.
$$
For a better idea of the precision of this calculation, we can rewrite eqref{ineq:2} as
$$
frac{489{,}609{,}908}{870{,}062{,}193}
< frac{1}{P} - 12 <
frac{4{,}840{,}699{,}348}{8{,}544{,}456{,}921},
$$
whence (this could be done by hand, although again I used a calculator)
$$
frac{1}{12.57} < P < frac{1}{12.56}.
$$
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5 Answers
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5 Answers
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Let $P_n:=prodlimits_{k=1}^n,dfrac{2k-1}{2k}$ for each integer $ngeq 1$. We shall prove that $$frac{2}{3sqrt{2n}}<P_n<frac{1}{sqrt{2n}}tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $dfrac{1}{sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$). The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2leq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k}{2k+1}right)=frac{1}{4},prod_{k=3}^{2n},frac{k}{k+1}=frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2geq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k-2}{2k-1}right)=frac{1}{4},prod_{k=2}^{2n-1},frac{k}{k+1}=frac{2}{4(2n)},.$$
This shows that
$$frac{1}{2,sqrt{n}}leq P_nleq frac{sqrt{3}}{2,sqrt{2n+1}},.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$. Note that this implies (*). In particular, for $n=50$, we have
$$frac{1}{15}<0.0707<frac{1}{2cdotsqrt{50}}<P_{50}<frac{sqrt{3}}{2cdot sqrt{101}}<0.0862<frac{1}{10},.$$
add a comment |
up vote
4
down vote
Let $P_n:=prodlimits_{k=1}^n,dfrac{2k-1}{2k}$ for each integer $ngeq 1$. We shall prove that $$frac{2}{3sqrt{2n}}<P_n<frac{1}{sqrt{2n}}tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $dfrac{1}{sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$). The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2leq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k}{2k+1}right)=frac{1}{4},prod_{k=3}^{2n},frac{k}{k+1}=frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2geq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k-2}{2k-1}right)=frac{1}{4},prod_{k=2}^{2n-1},frac{k}{k+1}=frac{2}{4(2n)},.$$
This shows that
$$frac{1}{2,sqrt{n}}leq P_nleq frac{sqrt{3}}{2,sqrt{2n+1}},.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$. Note that this implies (*). In particular, for $n=50$, we have
$$frac{1}{15}<0.0707<frac{1}{2cdotsqrt{50}}<P_{50}<frac{sqrt{3}}{2cdot sqrt{101}}<0.0862<frac{1}{10},.$$
add a comment |
up vote
4
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up vote
4
down vote
Let $P_n:=prodlimits_{k=1}^n,dfrac{2k-1}{2k}$ for each integer $ngeq 1$. We shall prove that $$frac{2}{3sqrt{2n}}<P_n<frac{1}{sqrt{2n}}tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $dfrac{1}{sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$). The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2leq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k}{2k+1}right)=frac{1}{4},prod_{k=3}^{2n},frac{k}{k+1}=frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2geq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k-2}{2k-1}right)=frac{1}{4},prod_{k=2}^{2n-1},frac{k}{k+1}=frac{2}{4(2n)},.$$
This shows that
$$frac{1}{2,sqrt{n}}leq P_nleq frac{sqrt{3}}{2,sqrt{2n+1}},.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$. Note that this implies (*). In particular, for $n=50$, we have
$$frac{1}{15}<0.0707<frac{1}{2cdotsqrt{50}}<P_{50}<frac{sqrt{3}}{2cdot sqrt{101}}<0.0862<frac{1}{10},.$$
Let $P_n:=prodlimits_{k=1}^n,dfrac{2k-1}{2k}$ for each integer $ngeq 1$. We shall prove that $$frac{2}{3sqrt{2n}}<P_n<frac{1}{sqrt{2n}}tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $dfrac{1}{sqrt{2}}$ there so that (*) implies the OP's inequality for $n=50$). The asymptotic behavior of $P_n$ is given here.
Observe that
$$P_n^2leq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k}{2k+1}right)=frac{1}{4},prod_{k=3}^{2n},frac{k}{k+1}=frac{3}{4(2n+1)}.$$
In addition,
$$P_n^2geq frac{1}{2^2},left(prod_{k=2}^n,frac{2k-1}{2k}right),left(prod_{k=2}^n,frac{2k-2}{2k-1}right)=frac{1}{4},prod_{k=2}^{2n-1},frac{k}{k+1}=frac{2}{4(2n)},.$$
This shows that
$$frac{1}{2,sqrt{n}}leq P_nleq frac{sqrt{3}}{2,sqrt{2n+1}},.$$
Both the inequality on the right and the inequality on the left have a unique equality case: $n=1$. Note that this implies (*). In particular, for $n=50$, we have
$$frac{1}{15}<0.0707<frac{1}{2cdotsqrt{50}}<P_{50}<frac{sqrt{3}}{2cdot sqrt{101}}<0.0862<frac{1}{10},.$$
edited yesterday
answered yesterday
Batominovski
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Let $S_n=prod_{k=1}^nfrac{2k-1}{2k}$. From Wallis' product, we have
$$prod_{k=1}^inftyleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)=frac{pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=1}^nleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)<frac{pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)prod_{k=1}^nleft(frac{2k-1}{2k}right)^2=prod_{k=1}^nleft(frac{2k-1}{2k}cdotfrac{2k+1}{2k}right)>frac{2}{pi}.$$
Therefore,
$$S_n>frac{1}{sqrt{pileft(n+frac12right)}}.$$
Similarly, Wallis' product also implies that
$$prod_{k=2}^inftyleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)=frac{4}{pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac{4}{pi}$$
for all $n$. That is,
$$2nS_n^2=2nprod_{k=1}^nleft(frac{2k-1}{2k}right)^2=frac12prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac12left(frac{4}{pi}right).$$
Therefore,
$$S_n<frac{1}{sqrt{pi n}}.$$ Hence,
$$frac{1}{sqrt{pileft(n+frac12right)}}<S_n<frac1{sqrt{pi n}}$$
for every $n$.
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Let $S_n=prod_{k=1}^nfrac{2k-1}{2k}$. From Wallis' product, we have
$$prod_{k=1}^inftyleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)=frac{pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=1}^nleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)<frac{pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)prod_{k=1}^nleft(frac{2k-1}{2k}right)^2=prod_{k=1}^nleft(frac{2k-1}{2k}cdotfrac{2k+1}{2k}right)>frac{2}{pi}.$$
Therefore,
$$S_n>frac{1}{sqrt{pileft(n+frac12right)}}.$$
Similarly, Wallis' product also implies that
$$prod_{k=2}^inftyleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)=frac{4}{pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac{4}{pi}$$
for all $n$. That is,
$$2nS_n^2=2nprod_{k=1}^nleft(frac{2k-1}{2k}right)^2=frac12prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac12left(frac{4}{pi}right).$$
Therefore,
$$S_n<frac{1}{sqrt{pi n}}.$$ Hence,
$$frac{1}{sqrt{pileft(n+frac12right)}}<S_n<frac1{sqrt{pi n}}$$
for every $n$.
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up vote
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down vote
Let $S_n=prod_{k=1}^nfrac{2k-1}{2k}$. From Wallis' product, we have
$$prod_{k=1}^inftyleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)=frac{pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=1}^nleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)<frac{pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)prod_{k=1}^nleft(frac{2k-1}{2k}right)^2=prod_{k=1}^nleft(frac{2k-1}{2k}cdotfrac{2k+1}{2k}right)>frac{2}{pi}.$$
Therefore,
$$S_n>frac{1}{sqrt{pileft(n+frac12right)}}.$$
Similarly, Wallis' product also implies that
$$prod_{k=2}^inftyleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)=frac{4}{pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac{4}{pi}$$
for all $n$. That is,
$$2nS_n^2=2nprod_{k=1}^nleft(frac{2k-1}{2k}right)^2=frac12prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac12left(frac{4}{pi}right).$$
Therefore,
$$S_n<frac{1}{sqrt{pi n}}.$$ Hence,
$$frac{1}{sqrt{pileft(n+frac12right)}}<S_n<frac1{sqrt{pi n}}$$
for every $n$.
Let $S_n=prod_{k=1}^nfrac{2k-1}{2k}$. From Wallis' product, we have
$$prod_{k=1}^inftyleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)=frac{pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=1}^nleft(frac{2k}{2k-1}cdotfrac{2k}{2k+1}right)<frac{pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)prod_{k=1}^nleft(frac{2k-1}{2k}right)^2=prod_{k=1}^nleft(frac{2k-1}{2k}cdotfrac{2k+1}{2k}right)>frac{2}{pi}.$$
Therefore,
$$S_n>frac{1}{sqrt{pileft(n+frac12right)}}.$$
Similarly, Wallis' product also implies that
$$prod_{k=2}^inftyleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)=frac{4}{pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac{4}{pi}$$
for all $n$. That is,
$$2nS_n^2=2nprod_{k=1}^nleft(frac{2k-1}{2k}right)^2=frac12prod_{k=2}^nleft(frac{2k-1}{2k-2}cdotfrac{2k-1}{2k}right)<frac12left(frac{4}{pi}right).$$
Therefore,
$$S_n<frac{1}{sqrt{pi n}}.$$ Hence,
$$frac{1}{sqrt{pileft(n+frac12right)}}<S_n<frac1{sqrt{pi n}}$$
for every $n$.
edited yesterday
answered yesterday
Zvi
3,230223
3,230223
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Let $a=frac{1}{2}.frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
and:
$b=frac{2}{3}.frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$frac{1}{2}<frac{2}{3}, frac{3}{4}<frac{4}{5}. . . frac{99}{100}<frac{100}{101}$
⇒ $a^2 < ab=(frac{1}{2}.frac{2}{3}).(frac{3}{4}.frac{4}{5}). . . .(frac{99}{100}.frac{100}{101})=frac{1}{101}$
$a^2<frac{1}{101}$⇒$a<frac{1}{sqrt{101}}<frac{1}{10}$
Also:
$2a=frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
$frac{3}{2}.b=frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
$2a<frac{3}{2}b$⇒$2a^2<frac{3}{2}ab=frac{3}{2}.frac{3}{101}$
Or $a^2<frac{3}{4}ab=frac{3}{4}.frac{3}{101}$
Since $9>4$ then $a^2 >frac{4}{9times 101}$ and therefore:
$a>frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=frac{1}{101}$⇒$a^2>frac{1}{202}$⇒$a>frac{1}{sqrt{202}}>frac{1}{sqrt{225}}=frac{1}{15}$
2
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
3
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
|
show 1 more comment
up vote
2
down vote
Let $a=frac{1}{2}.frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
and:
$b=frac{2}{3}.frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$frac{1}{2}<frac{2}{3}, frac{3}{4}<frac{4}{5}. . . frac{99}{100}<frac{100}{101}$
⇒ $a^2 < ab=(frac{1}{2}.frac{2}{3}).(frac{3}{4}.frac{4}{5}). . . .(frac{99}{100}.frac{100}{101})=frac{1}{101}$
$a^2<frac{1}{101}$⇒$a<frac{1}{sqrt{101}}<frac{1}{10}$
Also:
$2a=frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
$frac{3}{2}.b=frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
$2a<frac{3}{2}b$⇒$2a^2<frac{3}{2}ab=frac{3}{2}.frac{3}{101}$
Or $a^2<frac{3}{4}ab=frac{3}{4}.frac{3}{101}$
Since $9>4$ then $a^2 >frac{4}{9times 101}$ and therefore:
$a>frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=frac{1}{101}$⇒$a^2>frac{1}{202}$⇒$a>frac{1}{sqrt{202}}>frac{1}{sqrt{225}}=frac{1}{15}$
2
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
3
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
|
show 1 more comment
up vote
2
down vote
up vote
2
down vote
Let $a=frac{1}{2}.frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
and:
$b=frac{2}{3}.frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$frac{1}{2}<frac{2}{3}, frac{3}{4}<frac{4}{5}. . . frac{99}{100}<frac{100}{101}$
⇒ $a^2 < ab=(frac{1}{2}.frac{2}{3}).(frac{3}{4}.frac{4}{5}). . . .(frac{99}{100}.frac{100}{101})=frac{1}{101}$
$a^2<frac{1}{101}$⇒$a<frac{1}{sqrt{101}}<frac{1}{10}$
Also:
$2a=frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
$frac{3}{2}.b=frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
$2a<frac{3}{2}b$⇒$2a^2<frac{3}{2}ab=frac{3}{2}.frac{3}{101}$
Or $a^2<frac{3}{4}ab=frac{3}{4}.frac{3}{101}$
Since $9>4$ then $a^2 >frac{4}{9times 101}$ and therefore:
$a>frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=frac{1}{101}$⇒$a^2>frac{1}{202}$⇒$a>frac{1}{sqrt{202}}>frac{1}{sqrt{225}}=frac{1}{15}$
Let $a=frac{1}{2}.frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
and:
$b=frac{2}{3}.frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
It is clear that $a<b$ because each factor of $a$ is less than its corresponding factor in $b$ :
$frac{1}{2}<frac{2}{3}, frac{3}{4}<frac{4}{5}. . . frac{99}{100}<frac{100}{101}$
⇒ $a^2 < ab=(frac{1}{2}.frac{2}{3}).(frac{3}{4}.frac{4}{5}). . . .(frac{99}{100}.frac{100}{101})=frac{1}{101}$
$a^2<frac{1}{101}$⇒$a<frac{1}{sqrt{101}}<frac{1}{10}$
Also:
$2a=frac{3}{4}.frac{5}{6}. . . .frac{99}{100}$
$frac{3}{2}.b=frac{4}{5}.frac{6}{7}. . . frac{100}{101}$
$2a<frac{3}{2}b$⇒$2a^2<frac{3}{2}ab=frac{3}{2}.frac{3}{101}$
Or $a^2<frac{3}{4}ab=frac{3}{4}.frac{3}{101}$
Since $9>4$ then $a^2 >frac{4}{9times 101}$ and therefore:
$a>frac{1}{15}$
A more reliable reasoning is given as a comment for this part:
$2a>b$ ⇒ $2a^2>ab=frac{1}{101}$⇒$a^2>frac{1}{202}$⇒$a>frac{1}{sqrt{202}}>frac{1}{sqrt{225}}=frac{1}{15}$
edited 10 hours ago
answered yesterday
sirous
1,5331513
1,5331513
2
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
3
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
|
show 1 more comment
2
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
3
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
2
2
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
I've lost count of how many times I've read the last line, and I still can't understand: (i) how you get $frac{3}{2}ab=frac{3}{2}cdotfrac{3}{101}$, when surely it should be $frac{3}{2}ab=frac{3}{2}cdotfrac{1}{101}$; (ii) how you get from $2a^2<frac{3}{2}cdotfrac{3}{101}$ to $a^2>frac{4}{9times101}$, when surely it should be $frac{1}{a^2}>frac{4times101}{9}$; or (iii) generally, how a lower bound for $a$ could possibly be obtained from an upper bound for $a$! (My reading comprehension sometimes goes on the blink, so, in view of the upvotes, I guess it must have happened again!)
– Calum Gilhooley
yesterday
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
In the line $a^2<ab=...$ just eliminate the first term , because it is moved to LHS as 2a and (3/2)ab.
– sirous
23 hours ago
3
3
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@CalumGilhooley I have the same problem. But one could reason as follows: $$2a = frac{3}{4}cdot frac{5}{6} cdots frac{99}{100} color{blue}{cdot 1} > frac{2}{3}cdot frac{4}{5} cdots frac{100}{101} = b$$ $$Rightarrow 2a^2 > ab = frac{1}{101} Rightarrow a > frac{1}{sqrt{202}} > frac{1}{sqrt{225}} = frac{1}{15}$$
– trancelocation
21 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@sirous I imagine your reply, which I couldn't understand, was meant to be an explanation of $2a<frac{3}{2}b$, which wasn't any of the three things I was asking about, and which was one of the only two parts of the last line that I didn't need to have explained! In contrast, trancelocation's reply made perfect sense.
– Calum Gilhooley
20 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
@CalumGilhooley, I edited my answer for more clarification. The reasoning of -trancelocation is also good.
– sirous
13 hours ago
|
show 1 more comment
up vote
0
down vote
Just a thought, that may be worth mentioning:
The expression:
$$p=(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$frac{1.3.5.7}{2.4.6.8}=frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=frac{odd_numbers}{even_numbers}=frac{p_o}{p_e}=$$
$$p=frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<frac{(n!)}{((n!)2^{n})^2}$$
$$p<frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
add a comment |
up vote
0
down vote
Just a thought, that may be worth mentioning:
The expression:
$$p=(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$frac{1.3.5.7}{2.4.6.8}=frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=frac{odd_numbers}{even_numbers}=frac{p_o}{p_e}=$$
$$p=frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<frac{(n!)}{((n!)2^{n})^2}$$
$$p<frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
add a comment |
up vote
0
down vote
up vote
0
down vote
Just a thought, that may be worth mentioning:
The expression:
$$p=(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$frac{1.3.5.7}{2.4.6.8}=frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=frac{odd_numbers}{even_numbers}=frac{p_o}{p_e}=$$
$$p=frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<frac{(n!)}{((n!)2^{n})^2}$$
$$p<frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
Just a thought, that may be worth mentioning:
The expression:
$$p=(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})$$
We could use the following identities:
Product of $n$ odd numbers =
$$p_o = frac{(2n!)}{(n!)2^{n}}$$
Product of $n$ even numbers =
$$p_e = (n!)2^{n}$$
The first $4$ terms of $p$ =
$$frac{1.3.5.7}{2.4.6.8}=frac{105}{1152}=0.2734$$
We may write $p$ as:
$$p=frac{odd_numbers}{even_numbers}=frac{p_o}{p_e}=$$
$$p=frac{(2n)!}{((n!)2^{n})^2}$$
To prove $p<frac{1}{10}$, we know that (2n)! < n! (for n>1), so we can write:
$$p<frac{(n!)}{((n!)2^{n})^2}$$
$$p<frac{1}{(n!)({n})^2}$$
We could conclude that, for $n >=2$
$$p<frac{1}{10}$$
for the lower bound, any of the other solution presented may be considered or check stovf-math p between 1/13 and 1/15 may also be considered.
edited yesterday
answered yesterday
NoChance
3,58121221
3,58121221
add a comment |
add a comment |
up vote
0
down vote
This is tedious and unsophisticated, but don't knock it, it works! :)
As well as answering the present question, it also answers the linked question, prove $frac{1}{13}<frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}<frac{1}{12}$.
We need a preliminary lemma:
If
$$
k = left(1 - frac{1}{52^2}right)left(1 - frac{1}{56^2}right)cdotsleft(1 - frac{1}{96^2}right),
$$
then
$$
frac{383}{384} < k < frac{1300}{1303}.
$$
Rounding up and down, as appropriate, this is approximately
$$
0.997395 < k < 0.997698,
$$
but of course we avoid using such calculations.
Proof. By the Weierstrass product inequality, we have
$$
1 - s < k < frac{1}{1 + s},
$$
where
$$
s = frac{1}{52^2} + frac{1}{56^2} + cdots + frac{1}{96^2} =
frac{1}{16}left(frac{1}{13^2} + frac{1}{14^2} + cdots + frac{1}{24^2}right).
$$
Telescoping,
begin{align*}
16s & <
frac{1}{12cdot13} + frac{1}{13cdot14} + cdots + frac{1}{23cdot24}
= frac{1}{24}, \
16s & >
frac{1}{13cdot14} + frac{1}{14cdot15} + cdots + frac{1}{24cdot25}
= frac{12}{325},
end{align*}
therefore
$$
1 - frac{1}{384} < k < frac{1}{1 + frac{3}{1300}},
$$
as required. $square$
The number we wish to approximate is
$$
P = frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}
= frac{51cdot53cdot55cdots97cdot99}{2^{25}cdot4cdot8cdot12cdots96cdot100}
= kQ,
$$
where
begin{gather*}
Q = frac{52^2cdot56^2cdots96^2cdot99}{2^{75}cdot25!}
= frac{13cdot14cdots24cdot99}{2^{27}cdot12!cdot25}
= frac{13cdot15cdot17cdot19cdot21cdot23cdot99}{2^{21}cdot6!cdot25} \
= frac{7cdot13cdot17cdot19cdot23cdot99}{2^{25}cdot25}.
end{gather*}
Therefore, using the bounds obtained for $k$ in the lemma,
begin{equation}
tag{$1$}label{ineq:1}
frac{7cdot13cdot17cdot19cdot23cdot33cdot383}{2^{32}cdot25}
< P <
frac{7cdot13^2cdot17cdot19cdot23cdot99}{2^{23}cdot1303}.
end{equation}
Although this is well within the range of easy hand calculation, I lazily used a calculator for most of the work (only doubling $2^{22}cdot1303 = 5{,}465{,}178{,}112$ at the last stage by hand), to get
begin{equation}
tag{$2$}label{ineq:2}
frac{8{,}544{,}456{,}921}{107{,}374{,}182{,}400} < P < frac{870{,}062{,}193}{10{,}930{,}356{,}224}.
end{equation}
Approximately, rounding up and down again,
$$
0.079576 < P < 0.079601.
$$
One way to "simplify" eqref{ineq:1} (although it seems simpler to me to calculate eqref{ineq:2} and be done with it!) is as follows:
For the upper bound, observe that $7cdot13cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2cdot3cdot7cdot31$, so
$$
frac{7cdot13cdot19}{1303} < frac{1728}{2cdot3cdot7cdot31} = frac{288}{217} < frac{288}{216} = frac{4}{3},
$$
(as one can now easily verify with hindsight), therefore
$$
P < frac{13cdot17cdot23cdot99}{2^{21}cdot3} < frac{17cdot300cdot100}{2^{11}cdot3cdot1000} = frac{85}{1024} < frac{1}{12}.
$$
For the lower bound, first simplify $Q$ by noticing that $7cdot17cdot19cdot23 = 52003 > 52000 = 2^5cdot5^3cdot13$, whence
$$
Q > frac{5cdot13^2cdot99}{2^{20}} = frac{5cdot13cdot1287}{2^{20}} > frac{5cdot13cdot1280}{2^{20}} = frac{5^2cdot13}{2^{12}},
$$
and then, instead of using the lower bound from the lemma directly, weaken it to $P > frac{64}{65}Q$, which gives
$$
P > frac{5}{64} > frac{1}{13}.
$$
For a better idea of the precision of this calculation, we can rewrite eqref{ineq:2} as
$$
frac{489{,}609{,}908}{870{,}062{,}193}
< frac{1}{P} - 12 <
frac{4{,}840{,}699{,}348}{8{,}544{,}456{,}921},
$$
whence (this could be done by hand, although again I used a calculator)
$$
frac{1}{12.57} < P < frac{1}{12.56}.
$$
add a comment |
up vote
0
down vote
This is tedious and unsophisticated, but don't knock it, it works! :)
As well as answering the present question, it also answers the linked question, prove $frac{1}{13}<frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}<frac{1}{12}$.
We need a preliminary lemma:
If
$$
k = left(1 - frac{1}{52^2}right)left(1 - frac{1}{56^2}right)cdotsleft(1 - frac{1}{96^2}right),
$$
then
$$
frac{383}{384} < k < frac{1300}{1303}.
$$
Rounding up and down, as appropriate, this is approximately
$$
0.997395 < k < 0.997698,
$$
but of course we avoid using such calculations.
Proof. By the Weierstrass product inequality, we have
$$
1 - s < k < frac{1}{1 + s},
$$
where
$$
s = frac{1}{52^2} + frac{1}{56^2} + cdots + frac{1}{96^2} =
frac{1}{16}left(frac{1}{13^2} + frac{1}{14^2} + cdots + frac{1}{24^2}right).
$$
Telescoping,
begin{align*}
16s & <
frac{1}{12cdot13} + frac{1}{13cdot14} + cdots + frac{1}{23cdot24}
= frac{1}{24}, \
16s & >
frac{1}{13cdot14} + frac{1}{14cdot15} + cdots + frac{1}{24cdot25}
= frac{12}{325},
end{align*}
therefore
$$
1 - frac{1}{384} < k < frac{1}{1 + frac{3}{1300}},
$$
as required. $square$
The number we wish to approximate is
$$
P = frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}
= frac{51cdot53cdot55cdots97cdot99}{2^{25}cdot4cdot8cdot12cdots96cdot100}
= kQ,
$$
where
begin{gather*}
Q = frac{52^2cdot56^2cdots96^2cdot99}{2^{75}cdot25!}
= frac{13cdot14cdots24cdot99}{2^{27}cdot12!cdot25}
= frac{13cdot15cdot17cdot19cdot21cdot23cdot99}{2^{21}cdot6!cdot25} \
= frac{7cdot13cdot17cdot19cdot23cdot99}{2^{25}cdot25}.
end{gather*}
Therefore, using the bounds obtained for $k$ in the lemma,
begin{equation}
tag{$1$}label{ineq:1}
frac{7cdot13cdot17cdot19cdot23cdot33cdot383}{2^{32}cdot25}
< P <
frac{7cdot13^2cdot17cdot19cdot23cdot99}{2^{23}cdot1303}.
end{equation}
Although this is well within the range of easy hand calculation, I lazily used a calculator for most of the work (only doubling $2^{22}cdot1303 = 5{,}465{,}178{,}112$ at the last stage by hand), to get
begin{equation}
tag{$2$}label{ineq:2}
frac{8{,}544{,}456{,}921}{107{,}374{,}182{,}400} < P < frac{870{,}062{,}193}{10{,}930{,}356{,}224}.
end{equation}
Approximately, rounding up and down again,
$$
0.079576 < P < 0.079601.
$$
One way to "simplify" eqref{ineq:1} (although it seems simpler to me to calculate eqref{ineq:2} and be done with it!) is as follows:
For the upper bound, observe that $7cdot13cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2cdot3cdot7cdot31$, so
$$
frac{7cdot13cdot19}{1303} < frac{1728}{2cdot3cdot7cdot31} = frac{288}{217} < frac{288}{216} = frac{4}{3},
$$
(as one can now easily verify with hindsight), therefore
$$
P < frac{13cdot17cdot23cdot99}{2^{21}cdot3} < frac{17cdot300cdot100}{2^{11}cdot3cdot1000} = frac{85}{1024} < frac{1}{12}.
$$
For the lower bound, first simplify $Q$ by noticing that $7cdot17cdot19cdot23 = 52003 > 52000 = 2^5cdot5^3cdot13$, whence
$$
Q > frac{5cdot13^2cdot99}{2^{20}} = frac{5cdot13cdot1287}{2^{20}} > frac{5cdot13cdot1280}{2^{20}} = frac{5^2cdot13}{2^{12}},
$$
and then, instead of using the lower bound from the lemma directly, weaken it to $P > frac{64}{65}Q$, which gives
$$
P > frac{5}{64} > frac{1}{13}.
$$
For a better idea of the precision of this calculation, we can rewrite eqref{ineq:2} as
$$
frac{489{,}609{,}908}{870{,}062{,}193}
< frac{1}{P} - 12 <
frac{4{,}840{,}699{,}348}{8{,}544{,}456{,}921},
$$
whence (this could be done by hand, although again I used a calculator)
$$
frac{1}{12.57} < P < frac{1}{12.56}.
$$
add a comment |
up vote
0
down vote
up vote
0
down vote
This is tedious and unsophisticated, but don't knock it, it works! :)
As well as answering the present question, it also answers the linked question, prove $frac{1}{13}<frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}<frac{1}{12}$.
We need a preliminary lemma:
If
$$
k = left(1 - frac{1}{52^2}right)left(1 - frac{1}{56^2}right)cdotsleft(1 - frac{1}{96^2}right),
$$
then
$$
frac{383}{384} < k < frac{1300}{1303}.
$$
Rounding up and down, as appropriate, this is approximately
$$
0.997395 < k < 0.997698,
$$
but of course we avoid using such calculations.
Proof. By the Weierstrass product inequality, we have
$$
1 - s < k < frac{1}{1 + s},
$$
where
$$
s = frac{1}{52^2} + frac{1}{56^2} + cdots + frac{1}{96^2} =
frac{1}{16}left(frac{1}{13^2} + frac{1}{14^2} + cdots + frac{1}{24^2}right).
$$
Telescoping,
begin{align*}
16s & <
frac{1}{12cdot13} + frac{1}{13cdot14} + cdots + frac{1}{23cdot24}
= frac{1}{24}, \
16s & >
frac{1}{13cdot14} + frac{1}{14cdot15} + cdots + frac{1}{24cdot25}
= frac{12}{325},
end{align*}
therefore
$$
1 - frac{1}{384} < k < frac{1}{1 + frac{3}{1300}},
$$
as required. $square$
The number we wish to approximate is
$$
P = frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}
= frac{51cdot53cdot55cdots97cdot99}{2^{25}cdot4cdot8cdot12cdots96cdot100}
= kQ,
$$
where
begin{gather*}
Q = frac{52^2cdot56^2cdots96^2cdot99}{2^{75}cdot25!}
= frac{13cdot14cdots24cdot99}{2^{27}cdot12!cdot25}
= frac{13cdot15cdot17cdot19cdot21cdot23cdot99}{2^{21}cdot6!cdot25} \
= frac{7cdot13cdot17cdot19cdot23cdot99}{2^{25}cdot25}.
end{gather*}
Therefore, using the bounds obtained for $k$ in the lemma,
begin{equation}
tag{$1$}label{ineq:1}
frac{7cdot13cdot17cdot19cdot23cdot33cdot383}{2^{32}cdot25}
< P <
frac{7cdot13^2cdot17cdot19cdot23cdot99}{2^{23}cdot1303}.
end{equation}
Although this is well within the range of easy hand calculation, I lazily used a calculator for most of the work (only doubling $2^{22}cdot1303 = 5{,}465{,}178{,}112$ at the last stage by hand), to get
begin{equation}
tag{$2$}label{ineq:2}
frac{8{,}544{,}456{,}921}{107{,}374{,}182{,}400} < P < frac{870{,}062{,}193}{10{,}930{,}356{,}224}.
end{equation}
Approximately, rounding up and down again,
$$
0.079576 < P < 0.079601.
$$
One way to "simplify" eqref{ineq:1} (although it seems simpler to me to calculate eqref{ineq:2} and be done with it!) is as follows:
For the upper bound, observe that $7cdot13cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2cdot3cdot7cdot31$, so
$$
frac{7cdot13cdot19}{1303} < frac{1728}{2cdot3cdot7cdot31} = frac{288}{217} < frac{288}{216} = frac{4}{3},
$$
(as one can now easily verify with hindsight), therefore
$$
P < frac{13cdot17cdot23cdot99}{2^{21}cdot3} < frac{17cdot300cdot100}{2^{11}cdot3cdot1000} = frac{85}{1024} < frac{1}{12}.
$$
For the lower bound, first simplify $Q$ by noticing that $7cdot17cdot19cdot23 = 52003 > 52000 = 2^5cdot5^3cdot13$, whence
$$
Q > frac{5cdot13^2cdot99}{2^{20}} = frac{5cdot13cdot1287}{2^{20}} > frac{5cdot13cdot1280}{2^{20}} = frac{5^2cdot13}{2^{12}},
$$
and then, instead of using the lower bound from the lemma directly, weaken it to $P > frac{64}{65}Q$, which gives
$$
P > frac{5}{64} > frac{1}{13}.
$$
For a better idea of the precision of this calculation, we can rewrite eqref{ineq:2} as
$$
frac{489{,}609{,}908}{870{,}062{,}193}
< frac{1}{P} - 12 <
frac{4{,}840{,}699{,}348}{8{,}544{,}456{,}921},
$$
whence (this could be done by hand, although again I used a calculator)
$$
frac{1}{12.57} < P < frac{1}{12.56}.
$$
This is tedious and unsophisticated, but don't knock it, it works! :)
As well as answering the present question, it also answers the linked question, prove $frac{1}{13}<frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}<frac{1}{12}$.
We need a preliminary lemma:
If
$$
k = left(1 - frac{1}{52^2}right)left(1 - frac{1}{56^2}right)cdotsleft(1 - frac{1}{96^2}right),
$$
then
$$
frac{383}{384} < k < frac{1300}{1303}.
$$
Rounding up and down, as appropriate, this is approximately
$$
0.997395 < k < 0.997698,
$$
but of course we avoid using such calculations.
Proof. By the Weierstrass product inequality, we have
$$
1 - s < k < frac{1}{1 + s},
$$
where
$$
s = frac{1}{52^2} + frac{1}{56^2} + cdots + frac{1}{96^2} =
frac{1}{16}left(frac{1}{13^2} + frac{1}{14^2} + cdots + frac{1}{24^2}right).
$$
Telescoping,
begin{align*}
16s & <
frac{1}{12cdot13} + frac{1}{13cdot14} + cdots + frac{1}{23cdot24}
= frac{1}{24}, \
16s & >
frac{1}{13cdot14} + frac{1}{14cdot15} + cdots + frac{1}{24cdot25}
= frac{12}{325},
end{align*}
therefore
$$
1 - frac{1}{384} < k < frac{1}{1 + frac{3}{1300}},
$$
as required. $square$
The number we wish to approximate is
$$
P = frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdotsfrac{99}{100}
= frac{51cdot53cdot55cdots97cdot99}{2^{25}cdot4cdot8cdot12cdots96cdot100}
= kQ,
$$
where
begin{gather*}
Q = frac{52^2cdot56^2cdots96^2cdot99}{2^{75}cdot25!}
= frac{13cdot14cdots24cdot99}{2^{27}cdot12!cdot25}
= frac{13cdot15cdot17cdot19cdot21cdot23cdot99}{2^{21}cdot6!cdot25} \
= frac{7cdot13cdot17cdot19cdot23cdot99}{2^{25}cdot25}.
end{gather*}
Therefore, using the bounds obtained for $k$ in the lemma,
begin{equation}
tag{$1$}label{ineq:1}
frac{7cdot13cdot17cdot19cdot23cdot33cdot383}{2^{32}cdot25}
< P <
frac{7cdot13^2cdot17cdot19cdot23cdot99}{2^{23}cdot1303}.
end{equation}
Although this is well within the range of easy hand calculation, I lazily used a calculator for most of the work (only doubling $2^{22}cdot1303 = 5{,}465{,}178{,}112$ at the last stage by hand), to get
begin{equation}
tag{$2$}label{ineq:2}
frac{8{,}544{,}456{,}921}{107{,}374{,}182{,}400} < P < frac{870{,}062{,}193}{10{,}930{,}356{,}224}.
end{equation}
Approximately, rounding up and down again,
$$
0.079576 < P < 0.079601.
$$
One way to "simplify" eqref{ineq:1} (although it seems simpler to me to calculate eqref{ineq:2} and be done with it!) is as follows:
For the upper bound, observe that $7cdot13cdot19 = 1729 = 1 + 12^3$ (famously!), and $1302 = 2cdot3cdot7cdot31$, so
$$
frac{7cdot13cdot19}{1303} < frac{1728}{2cdot3cdot7cdot31} = frac{288}{217} < frac{288}{216} = frac{4}{3},
$$
(as one can now easily verify with hindsight), therefore
$$
P < frac{13cdot17cdot23cdot99}{2^{21}cdot3} < frac{17cdot300cdot100}{2^{11}cdot3cdot1000} = frac{85}{1024} < frac{1}{12}.
$$
For the lower bound, first simplify $Q$ by noticing that $7cdot17cdot19cdot23 = 52003 > 52000 = 2^5cdot5^3cdot13$, whence
$$
Q > frac{5cdot13^2cdot99}{2^{20}} = frac{5cdot13cdot1287}{2^{20}} > frac{5cdot13cdot1280}{2^{20}} = frac{5^2cdot13}{2^{12}},
$$
and then, instead of using the lower bound from the lemma directly, weaken it to $P > frac{64}{65}Q$, which gives
$$
P > frac{5}{64} > frac{1}{13}.
$$
For a better idea of the precision of this calculation, we can rewrite eqref{ineq:2} as
$$
frac{489{,}609{,}908}{870{,}062{,}193}
< frac{1}{P} - 12 <
frac{4{,}840{,}699{,}348}{8{,}544{,}456{,}921},
$$
whence (this could be done by hand, although again I used a calculator)
$$
frac{1}{12.57} < P < frac{1}{12.56}.
$$
edited 11 hours ago
answered yesterday
Calum Gilhooley
3,982529
3,982529
add a comment |
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3
Related
– Kemono Chen
yesterday
2
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
yesterday
2
@KemonoChen I think the correct inequality should be $$frac{2}{3sqrt{2n}}<frac12cdotfrac34cdotldotscdotfrac{2n-1}{2n}<frac{1}{sqrt{2n}},.$$ Otherwise, the left-hand side inequality is not even true for $n=1$ (and it is never true), while the right-hand side inequality is very weak.
– Batominovski
yesterday
3
$$frac{1}{4^n}binom{2n}{n}approx frac{1}{sqrt{pileft(n+frac{1}{4}right)}}.$$
– Jack D'Aurizio
yesterday
2
I found that $$frac{1}{sqrt{pi left(n+frac12right)}}<prod_{k=1}^nleft(frac{2k-1}{2k}right)<frac{1}{sqrt{pi n}}$$ with @JackD'Aurizio's estimate being extremely close. For $n=50$, the required product is between $0.07939248$ and $0.07979788$, with Jack's estimate being $0.07958972$. The actual value is $0.07958923$.
– Zvi
yesterday