Limit of the sum and Cauchy theorem on limits
If a sequence
${x_n}$
converges to l, then the sequence
${y_n}$
also converges to l,
where $y_n=x_1+x_2+cdots +dfrac{x_n}{n}$.
I suspect that $lim_{ntoinfty}sum_{k=1}^n dfrac{2k-1}{n^2}$ is 2, using Cauchy theorem, but the answer is 1. Please correct me if I'm wrong.
My computation
real-analysis limits
edited 2 days ago
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
add a comment |
If a sequence
${x_n}$
converges to l, then the sequence
${y_n}$
also converges to l,
where $y_n=x_1+x_2+cdots +dfrac{x_n}{n}$.
I suspect that $lim_{ntoinfty}sum_{k=1}^n dfrac{2k-1}{n^2}$ is 2, using Cauchy theorem, but the answer is 1. Please correct me if I'm wrong.
My computation
real-analysis limits
edited 2 days ago
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
– Robert Z
May 19 '18 at 5:47
Yes I meant Cauchy theorem for limits
– Josh
May 19 '18 at 6:23
Why my computation is wrong?
– Josh
May 19 '18 at 6:30
@Josh Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
May 31 '18 at 22:09
add a comment |
If a sequence
${x_n}$
converges to l, then the sequence
${y_n}$
also converges to l,
where $y_n=x_1+x_2+cdots +dfrac{x_n}{n}$.
I suspect that $lim_{ntoinfty}sum_{k=1}^n dfrac{2k-1}{n^2}$ is 2, using Cauchy theorem, but the answer is 1. Please correct me if I'm wrong.
My computation
real-analysis limits
edited 2 days ago
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
If a sequence
${x_n}$
converges to l, then the sequence
${y_n}$
also converges to l,
where $y_n=x_1+x_2+cdots +dfrac{x_n}{n}$.
I suspect that $lim_{ntoinfty}sum_{k=1}^n dfrac{2k-1}{n^2}$ is 2, using Cauchy theorem, but the answer is 1. Please correct me if I'm wrong.
My computation
real-analysis limits
real-analysis limits
edited 2 days ago
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
edited 2 days ago
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
edited 2 days ago
amWhy
192k28225439
edited 2 days ago
amWhy
192k28225439
edited 2 days ago
amWhy
192k28225439
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
asked May 19 '18 at 4:42
JoshJosh
62
asked May 19 '18 at 4:42
JoshJosh
62
62
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
– Robert Z
May 19 '18 at 5:47
Yes I meant Cauchy theorem for limits
– Josh
May 19 '18 at 6:23
Why my computation is wrong?
– Josh
May 19 '18 at 6:30
@Josh Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
May 31 '18 at 22:09
add a comment |
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
– Robert Z
May 19 '18 at 5:47
Yes I meant Cauchy theorem for limits
– Josh
May 19 '18 at 6:23
Why my computation is wrong?
– Josh
May 19 '18 at 6:30
@Josh Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
May 31 '18 at 22:09
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
– Robert Z
May 19 '18 at 5:47
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
– Robert Z
May 19 '18 at 5:47
Yes I meant Cauchy theorem for limits
– Josh
May 19 '18 at 6:23
Yes I meant Cauchy theorem for limits
– Josh
May 19 '18 at 6:23
Why my computation is wrong?
– Josh
May 19 '18 at 6:30
Why my computation is wrong?
– Josh
May 19 '18 at 6:30
@Josh Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
May 31 '18 at 22:09
@Josh Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
May 31 '18 at 22:09
add a comment |
1 Answer
1
active
oldest
votes
HINT
Note that
$$sum_{k=1}^n frac{2k-1}{n^2}=frac2{n^2}sum_{k=1}^n k-frac1{n^2}sum_{k=1}^n 1=frac2{n^2}frac{n(n+1)}{2}-frac1n$$
The first claim is true for
$$y_n=frac{x_1+x_2+⋯+x_n}{n}to l$$
by Stolz-Cesaro.
answered May 19 '18 at 6:06
gimusigimusi
1
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
Note that
$$sum_{k=1}^n frac{2k-1}{n^2}=frac2{n^2}sum_{k=1}^n k-frac1{n^2}sum_{k=1}^n 1=frac2{n^2}frac{n(n+1)}{2}-frac1n$$
The first claim is true for
$$y_n=frac{x_1+x_2+⋯+x_n}{n}to l$$
by Stolz-Cesaro.
answered May 19 '18 at 6:06
gimusigimusi
1
add a comment |
HINT
Note that
$$sum_{k=1}^n frac{2k-1}{n^2}=frac2{n^2}sum_{k=1}^n k-frac1{n^2}sum_{k=1}^n 1=frac2{n^2}frac{n(n+1)}{2}-frac1n$$
The first claim is true for
$$y_n=frac{x_1+x_2+⋯+x_n}{n}to l$$
by Stolz-Cesaro.
answered May 19 '18 at 6:06
gimusigimusi
1
add a comment |
HINT
Note that
$$sum_{k=1}^n frac{2k-1}{n^2}=frac2{n^2}sum_{k=1}^n k-frac1{n^2}sum_{k=1}^n 1=frac2{n^2}frac{n(n+1)}{2}-frac1n$$
The first claim is true for
$$y_n=frac{x_1+x_2+⋯+x_n}{n}to l$$
by Stolz-Cesaro.
answered May 19 '18 at 6:06
gimusigimusi
1
HINT
Note that
$$sum_{k=1}^n frac{2k-1}{n^2}=frac2{n^2}sum_{k=1}^n k-frac1{n^2}sum_{k=1}^n 1=frac2{n^2}frac{n(n+1)}{2}-frac1n$$
The first claim is true for
$$y_n=frac{x_1+x_2+⋯+x_n}{n}to l$$
by Stolz-Cesaro.
answered May 19 '18 at 6:06
gimusigimusi
1
answered May 19 '18 at 6:06
gimusigimusi
1
answered May 19 '18 at 6:06
gimusigimusi
1
answered May 19 '18 at 6:06
gimusigimusi
1
1
add a comment |
add a comment |
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Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
– Robert Z
May 19 '18 at 5:47
Yes I meant Cauchy theorem for limits
– Josh
May 19 '18 at 6:23
Why my computation is wrong?
– Josh
May 19 '18 at 6:30
@Josh Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
May 31 '18 at 22:09