Mixing
Limit of the sum and Cauchy theorem on limits
$begingroup$
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 Jan 5 at 16:19
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 16:19
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
$endgroup$
$begingroup$
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
$endgroup$
– Robert Z
May 19 '18 at 5:47
$begingroup$
Yes I meant Cauchy theorem for limits
$endgroup$
– Josh
May 19 '18 at 6:23
$begingroup$
Why my computation is wrong?
$endgroup$
– Josh
May 19 '18 at 6:30
$begingroup$
@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/…
$endgroup$
– gimusi
May 31 '18 at 22:09
add a comment |
$begingroup$
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 Jan 5 at 16:19
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
$endgroup$
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 Jan 5 at 16:19
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
edited Jan 5 at 16:19
amWhy
192k28225439
asked May 19 '18 at 4:42
JoshJosh
62
edited Jan 5 at 16:19
amWhy
192k28225439
edited Jan 5 at 16:19
amWhy
192k28225439
edited Jan 5 at 16:19
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
$begingroup$
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
$endgroup$
– Robert Z
May 19 '18 at 5:47
$begingroup$
Yes I meant Cauchy theorem for limits
$endgroup$
– Josh
May 19 '18 at 6:23
$begingroup$
Why my computation is wrong?
$endgroup$
– Josh
May 19 '18 at 6:30
$begingroup$
@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/…
$endgroup$
– gimusi
May 31 '18 at 22:09
add a comment |
$begingroup$
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
$endgroup$
– Robert Z
May 19 '18 at 5:47
$begingroup$
Yes I meant Cauchy theorem for limits
$endgroup$
– Josh
May 19 '18 at 6:23
$begingroup$
Why my computation is wrong?
$endgroup$
– Josh
May 19 '18 at 6:30
$begingroup$
@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/…
$endgroup$
– gimusi
May 31 '18 at 22:09
$begingroup$
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
$endgroup$
– Robert Z
May 19 '18 at 5:47
$begingroup$
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
$endgroup$
– Robert Z
May 19 '18 at 5:47
$begingroup$
Yes I meant Cauchy theorem for limits
$endgroup$
– Josh
May 19 '18 at 6:23
$begingroup$
Yes I meant Cauchy theorem for limits
$endgroup$
– Josh
May 19 '18 at 6:23
$begingroup$
Why my computation is wrong?
$endgroup$
– Josh
May 19 '18 at 6:30
$begingroup$
Why my computation is wrong?
$endgroup$
– Josh
May 19 '18 at 6:30
$begingroup$
@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/…
$endgroup$
– gimusi
May 31 '18 at 22:09
$begingroup$
@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/…
$endgroup$
– gimusi
May 31 '18 at 22:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
92.9k94494
$endgroup$
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
$begingroup$
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
92.9k94494
$endgroup$
add a comment |
$begingroup$
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
92.9k94494
$endgroup$
add a comment |
$begingroup$
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
92.9k94494
$endgroup$
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
92.9k94494
answered May 19 '18 at 6:06
gimusigimusi
92.9k94494
answered May 19 '18 at 6:06
gimusigimusi
92.9k94494
answered May 19 '18 at 6:06
gimusigimusi
92.9k94494
92.9k94494
add a comment |
add a comment |
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$begingroup$
Note that $sum_{k=1}^n (2k-1)=n^2$. Do you mean $y_n=frac{x_1+x_2+⋯+x_n}{n}$?
$endgroup$
– Robert Z
May 19 '18 at 5:47
$begingroup$
Yes I meant Cauchy theorem for limits
$endgroup$
– Josh
May 19 '18 at 6:23
$begingroup$
Why my computation is wrong?
$endgroup$
– Josh
May 19 '18 at 6:30
$begingroup$
@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/…
$endgroup$
– gimusi
May 31 '18 at 22:09