How to show $lim_{ntoinfty}nleft{sum_{k=1}^nfrac{1}{(n+k)^2}right}=frac{1}{2}$
Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$
Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.
Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$
That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is
... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.
real-analysis limits summation riemann-sum
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
asked Dec 5 '18 at 10:47
kaisakaisa
1019
add a comment |
Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$
Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.
Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$
That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is
... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.
real-analysis limits summation riemann-sum
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
asked Dec 5 '18 at 10:47
kaisakaisa
1019
6
Ah yes, the famous Corollary 8.3. I love that corollary
– mathworker21
Dec 5 '18 at 10:49
3
Should we all know what Corollary 8.3 is?
– José Carlos Santos
Dec 5 '18 at 10:49
@mathworker21 Ooops. I added it now. Thank you.
– kaisa
Dec 5 '18 at 10:58
See math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Dec 5 '18 at 10:59
add a comment |
Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$
Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.
Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$
That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is
... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.
real-analysis limits summation riemann-sum
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
asked Dec 5 '18 at 10:47
kaisakaisa
1019
Show that $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=dfrac{1}{2}.$$
Proof:
We can rewrite $$lim_{ntoinfty}nBigg{dfrac{1}{(n+1)^2}+dfrac{1}{(n+2)^2}+dfrac{1}{(n+3)^2}+cdots+dfrac{1}{(2n)^2}Bigg}=lim_{ntoinfty}nBigg{sum_{k=1}^ndfrac{1}{(n+k)^2}Bigg}$$
Which looks astoundingly similar to the form in Corollary 8.3, where $dfrac{b-a}{n} = nimplies b-a = 2n$.
Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$int_a^bf(x)dx=lim_{ntoinfty}dfrac{b-a}{n}sum_{k=0}^{n-1}fBigg(a+dfrac{k}{n}(b-a)Bigg)$$
That would mean the term in Corollary 8.3 would morph into
$$fleft(a+dfrac{k}{n}(b-a)right)=fleft((b-2n)+dfrac{k}{n}(2n)right)=f(b-2n+2k)$$
We need $f(b-2n+2k)$ to look like $dfrac{1}{(n+k)^2}$. This would imply that $f$ is
... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.
real-analysis limits summation riemann-sum
real-analysis limits summation riemann-sum
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
asked Dec 5 '18 at 10:47
kaisakaisa
1019
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
asked Dec 5 '18 at 10:47
kaisakaisa
1019
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
edited Dec 5 '18 at 11:13
Lorenzo B.
1,8402520
1,8402520
asked Dec 5 '18 at 10:47
kaisakaisa
1019
asked Dec 5 '18 at 10:47
kaisakaisa
1019
asked Dec 5 '18 at 10:47
kaisakaisa
1019
1019
6
Ah yes, the famous Corollary 8.3. I love that corollary
– mathworker21
Dec 5 '18 at 10:49
3
Should we all know what Corollary 8.3 is?
– José Carlos Santos
Dec 5 '18 at 10:49
@mathworker21 Ooops. I added it now. Thank you.
– kaisa
Dec 5 '18 at 10:58
See math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Dec 5 '18 at 10:59
add a comment |
6
Ah yes, the famous Corollary 8.3. I love that corollary
– mathworker21
Dec 5 '18 at 10:49
3
Should we all know what Corollary 8.3 is?
– José Carlos Santos
Dec 5 '18 at 10:49
@mathworker21 Ooops. I added it now. Thank you.
– kaisa
Dec 5 '18 at 10:58
See math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Dec 5 '18 at 10:59
6
6
Ah yes, the famous Corollary 8.3. I love that corollary
– mathworker21
Dec 5 '18 at 10:49
Ah yes, the famous Corollary 8.3. I love that corollary
– mathworker21
Dec 5 '18 at 10:49
3
3
Should we all know what Corollary 8.3 is?
– José Carlos Santos
Dec 5 '18 at 10:49
Should we all know what Corollary 8.3 is?
– José Carlos Santos
Dec 5 '18 at 10:49
@mathworker21 Ooops. I added it now. Thank you.
– kaisa
Dec 5 '18 at 10:58
@mathworker21 Ooops. I added it now. Thank you.
– kaisa
Dec 5 '18 at 10:58
See math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Dec 5 '18 at 10:59
See math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Dec 5 '18 at 10:59
add a comment |
2 Answers
2
active
oldest
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HINT
By Riemann's sum
$$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$
Refer also to the related
- Perfect understanding of Riemann Sums
answered Dec 5 '18 at 10:52
gimusigimusi
1
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
add a comment |
Another approach is to consider that the same result is obtained by the series
$$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$
Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.
And then to use
$$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$
Which simplifies the summation ...
answered Dec 5 '18 at 11:24
DamienDamien
52714
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
By Riemann's sum
$$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$
Refer also to the related
- Perfect understanding of Riemann Sums
answered Dec 5 '18 at 10:52
gimusigimusi
1
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
add a comment |
HINT
By Riemann's sum
$$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$
Refer also to the related
- Perfect understanding of Riemann Sums
answered Dec 5 '18 at 10:52
gimusigimusi
1
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
add a comment |
HINT
By Riemann's sum
$$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$
Refer also to the related
- Perfect understanding of Riemann Sums
answered Dec 5 '18 at 10:52
gimusigimusi
1
HINT
By Riemann's sum
$$lim_{ntoinfty}nsum_{k=1}^ndfrac{1}{(n+k)^2}=lim_{ntoinfty}frac1nsum_{k=1}^ndfrac{1}{(1+k/n)^2}$$
Refer also to the related
- Perfect understanding of Riemann Sums
answered Dec 5 '18 at 10:52
gimusigimusi
1
answered Dec 5 '18 at 10:52
gimusigimusi
1
answered Dec 5 '18 at 10:52
gimusigimusi
1
answered Dec 5 '18 at 10:52
gimusigimusi
1
1
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
add a comment |
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
But then I get to here: $lim_{ntoinfty}dfrac{n}{n}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}=lim_{ntoinfty}Bigg{sum_{k=1}^ndfrac{1}{(1+(k/n))^2}Bigg}$, which is infinity. EDIT: Ah I see. I forgot to square.
– kaisa
Dec 5 '18 at 11:09
add a comment |
Another approach is to consider that the same result is obtained by the series
$$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$
Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.
And then to use
$$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$
Which simplifies the summation ...
answered Dec 5 '18 at 11:24
DamienDamien
52714
add a comment |
Another approach is to consider that the same result is obtained by the series
$$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$
Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.
And then to use
$$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$
Which simplifies the summation ...
answered Dec 5 '18 at 11:24
DamienDamien
52714
add a comment |
Another approach is to consider that the same result is obtained by the series
$$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$
Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.
And then to use
$$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$
Which simplifies the summation ...
answered Dec 5 '18 at 11:24
DamienDamien
52714
Another approach is to consider that the same result is obtained by the series
$$lim_{ntoinfty}{sum_{k=1}^ndfrac{n}{(n+k)(n+k+1)}}$$
Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.
And then to use
$$frac{1}{(n+k)(n+k+1)} = frac{1}{n+k} - frac{1}{n+k+1} $$
Which simplifies the summation ...
answered Dec 5 '18 at 11:24
DamienDamien
52714
answered Dec 5 '18 at 11:24
DamienDamien
52714
answered Dec 5 '18 at 11:24
DamienDamien
52714
answered Dec 5 '18 at 11:24
DamienDamien
52714
52714
add a comment |
add a comment |
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6
Ah yes, the famous Corollary 8.3. I love that corollary
– mathworker21
Dec 5 '18 at 10:49
3
Should we all know what Corollary 8.3 is?
– José Carlos Santos
Dec 5 '18 at 10:49
@mathworker21 Ooops. I added it now. Thank you.
– kaisa
Dec 5 '18 at 10:58
See math.stackexchange.com/questions/469885/…
– lab bhattacharjee
Dec 5 '18 at 10:59