Show the value of the series: $sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$
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So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
add a comment |
up vote
-1
down vote
favorite
So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type$(-1)^{frac{n(n + 1)}{2}}$
.
– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
So i have to show the value of the series
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$
For this to exist it has to be convergent, so i checked if it is divergent using the limit test:
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$
Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?
Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n
real-analysis sequences-and-series limits analysis
real-analysis sequences-and-series limits analysis
edited Nov 22 at 10:38
gimusi
87.4k74393
87.4k74393
asked Nov 22 at 10:23
SirHawrk
11
11
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type$(-1)^{frac{n(n + 1)}{2}}$
.
– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
add a comment |
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type$(-1)^{frac{n(n + 1)}{2}}$
.
– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type
$(-1)^{frac{n(n + 1)}{2}}$
.– N. F. Taussig
Nov 22 at 10:35
There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type
$(-1)^{frac{n(n + 1)}{2}}$
.– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
up vote
0
down vote
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
up vote
1
down vote
up vote
1
down vote
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.
To evaluate the sum note that
- $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$
then for $n=4k$
$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$
With reference to the series note that
$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$
therefore we can't conclude form here that the series diverges.
Otherwise the series clearly converges since
$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$
and the RHS is a convergent geometric series.
edited Nov 22 at 11:39
answered Nov 22 at 10:31
gimusi
87.4k74393
87.4k74393
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
$(-1)^{i(i+1)/2}$ is $1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1$ for $i$ from $0$ to $10$. So it does not follow your even/odd rule. It does have period $4$, so try doing four sums, based on $ipmod{4}$.
– GEdgar
Nov 22 at 10:57
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
@GEdgar Opssss...thanks I fix that!
– gimusi
Nov 22 at 11:29
add a comment |
up vote
0
down vote
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
add a comment |
up vote
0
down vote
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
add a comment |
up vote
0
down vote
up vote
0
down vote
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Form $S$ in the following way:
$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$
Separate the sum accordance with even and odd $k$:
If k even then $k=2m$, if odd then $k=2m+1$
Substitute them into S we get:
$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$
Repeat the separation for m values:
If m even then $k=2r$, if odd then $k=2r+1$
we have:
$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$
Summing the geometrical series and perform the simplifications we get that:
$S=frac{9}{10}$
Using this method you can easily determine the value sum from k=0 to n.
answered Nov 22 at 17:30
JV.Stalker
46639
46639
add a comment |
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There is a close parenthesis missing from your expression. If you wish to type $(-1)^{frac{n(n + 1)}{2}}$, type
$(-1)^{frac{n(n + 1)}{2}}$
.– N. F. Taussig
Nov 22 at 10:35
This is a sum not a series, its limit as $nto infty$ is a series. Please clarify that.
– gimusi
Nov 22 at 10:42
Also the limit is wrong, indeed $(1/3)^n to 0$.
– gimusi
Nov 22 at 10:43