Mixing
$lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c , c > 0$ . Then, is ${x_n/n}$ convergent?
$begingroup$
If ${x_n}$ is a sequence which satisfies $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of ${x_n/n}$.
Attempt: $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c >0$
=> $x_n$ is unbounded and divergent.
However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
real-analysis sequences-and-series
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
$endgroup$
add a comment |
$begingroup$
If ${x_n}$ is a sequence which satisfies $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of ${x_n/n}$.
Attempt: $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c >0$
=> $x_n$ is unbounded and divergent.
However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
real-analysis sequences-and-series
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
$endgroup$
3
$begingroup$
Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=sum_{k=1}^n a_k$. So, the assumption is that $a_nto c$ and we are looking for the limit of the sequence of averages of $a_k$'s.
$endgroup$
– Berci
Feb 9 '14 at 18:27
2
$begingroup$
This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one.
$endgroup$
– Martin Sleziak
May 31 '14 at 12:01
add a comment |
$begingroup$
If ${x_n}$ is a sequence which satisfies $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of ${x_n/n}$.
Attempt: $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c >0$
=> $x_n$ is unbounded and divergent.
However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
real-analysis sequences-and-series
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
$endgroup$
If ${x_n}$ is a sequence which satisfies $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of ${x_n/n}$.
Attempt: $lim_{n rightarrowinfty} ~ x_{n+1} - x_n= c$ where $c >0$
=> $x_n$ is unbounded and divergent.
However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
edited May 31 '14 at 11:58
Martin Sleziak
44.7k9117272
44.7k9117272
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
asked Feb 9 '14 at 18:19
MathManMathMan
3,64841870
3,64841870
3
$begingroup$
Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=sum_{k=1}^n a_k$. So, the assumption is that $a_nto c$ and we are looking for the limit of the sequence of averages of $a_k$'s.
$endgroup$
– Berci
Feb 9 '14 at 18:27
2
$begingroup$
This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one.
$endgroup$
– Martin Sleziak
May 31 '14 at 12:01
add a comment |
3
$begingroup$
Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=sum_{k=1}^n a_k$. So, the assumption is that $a_nto c$ and we are looking for the limit of the sequence of averages of $a_k$'s.
$endgroup$
– Berci
Feb 9 '14 at 18:27
2
$begingroup$
This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one.
$endgroup$
– Martin Sleziak
May 31 '14 at 12:01
3
3
$begingroup$
Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=sum_{k=1}^n a_k$. So, the assumption is that $a_nto c$ and we are looking for the limit of the sequence of averages of $a_k$'s.
$endgroup$
– Berci
Feb 9 '14 at 18:27
$begingroup$
Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=sum_{k=1}^n a_k$. So, the assumption is that $a_nto c$ and we are looking for the limit of the sequence of averages of $a_k$'s.
$endgroup$
– Berci
Feb 9 '14 at 18:27
2
2
$begingroup$
This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one.
$endgroup$
– Martin Sleziak
May 31 '14 at 12:01
$begingroup$
This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one.
$endgroup$
– Martin Sleziak
May 31 '14 at 12:01
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We claim that $x_n/nrightarrow c$.
As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=sum_{k=1}^n a_k$ $color{red} { (1) }$. The assumption becomes $a_nrightarrow c$ and $x_n/n=frac{1}{n}sum_{k=1}^n a_k$.
Then
$$x_n/n-c=left(frac{1}{n}sum_{k=1}^n a_kright) - c = frac{1}{n} sum_{k=1}^n (a_k-c)quad color{red} { (2) } $$
Let $epsilon>0$ and choose $N$ large enough such that $|a_n-c|<epsilon$ for all $nge N$. Then
$$left|frac{x_n}{n}-cright|le frac{1}{n}sum_{k=1}^N |a_k-c| + frac{n-N}{n}epsilon$$
for $nge N$. Now let $nrightarrowinfty$ while keeping $N,epsilon$ fixed. Then we obtain
$$limsup_{nrightarrowinfty}left|frac{x_n}{n}-cright|le epsilon$$
Since $epsilon$ was arbitrary, the conclusion follows.
Note:
In essence we just proved that a summable series is also Cesàro summable.
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
$endgroup$
add a comment |
$begingroup$
That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$
$$lim_{n to infty} x_{n+1}-x_n = lim_{n to infty} frac{x_n}{n}$$
when $lim_{n to infty} x_{n+1}-x_n$ exists.
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
$endgroup$
add a comment |
$begingroup$
Apologies for editing Your Ad Here, but can someone please check (1) and (2)?
(1) = $sum_{1 le k le n} x_k - x_{k - 1}$ which telecopes to $-x_0 + x_n$, not just $x_n$ ?
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
$endgroup$
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
add a comment |
$begingroup$
We let $a_n:=x_{n+1}-x_n$ and use this theorem to conclude that $$frac{a_1+a_2+...+a_n}{n}to c implies frac{(x_2-x_1)+(x_3-x_2)+....+(x_{n+1}-x_n)}{n}to c $$ $$ implies frac{x_{n+1}-x_1}{n}to c implies frac{x_n}{n}to c$$ since $x_1$ is finite.
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
$endgroup$
add a comment |
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4 Answers
4
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4 Answers
4
active
oldest
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active
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active
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votes
$begingroup$
We claim that $x_n/nrightarrow c$.
As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=sum_{k=1}^n a_k$ $color{red} { (1) }$. The assumption becomes $a_nrightarrow c$ and $x_n/n=frac{1}{n}sum_{k=1}^n a_k$.
Then
$$x_n/n-c=left(frac{1}{n}sum_{k=1}^n a_kright) - c = frac{1}{n} sum_{k=1}^n (a_k-c)quad color{red} { (2) } $$
Let $epsilon>0$ and choose $N$ large enough such that $|a_n-c|<epsilon$ for all $nge N$. Then
$$left|frac{x_n}{n}-cright|le frac{1}{n}sum_{k=1}^N |a_k-c| + frac{n-N}{n}epsilon$$
for $nge N$. Now let $nrightarrowinfty$ while keeping $N,epsilon$ fixed. Then we obtain
$$limsup_{nrightarrowinfty}left|frac{x_n}{n}-cright|le epsilon$$
Since $epsilon$ was arbitrary, the conclusion follows.
Note:
In essence we just proved that a summable series is also Cesàro summable.
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
$endgroup$
add a comment |
$begingroup$
We claim that $x_n/nrightarrow c$.
As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=sum_{k=1}^n a_k$ $color{red} { (1) }$. The assumption becomes $a_nrightarrow c$ and $x_n/n=frac{1}{n}sum_{k=1}^n a_k$.
Then
$$x_n/n-c=left(frac{1}{n}sum_{k=1}^n a_kright) - c = frac{1}{n} sum_{k=1}^n (a_k-c)quad color{red} { (2) } $$
Let $epsilon>0$ and choose $N$ large enough such that $|a_n-c|<epsilon$ for all $nge N$. Then
$$left|frac{x_n}{n}-cright|le frac{1}{n}sum_{k=1}^N |a_k-c| + frac{n-N}{n}epsilon$$
for $nge N$. Now let $nrightarrowinfty$ while keeping $N,epsilon$ fixed. Then we obtain
$$limsup_{nrightarrowinfty}left|frac{x_n}{n}-cright|le epsilon$$
Since $epsilon$ was arbitrary, the conclusion follows.
Note:
In essence we just proved that a summable series is also Cesàro summable.
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
$endgroup$
add a comment |
$begingroup$
We claim that $x_n/nrightarrow c$.
As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=sum_{k=1}^n a_k$ $color{red} { (1) }$. The assumption becomes $a_nrightarrow c$ and $x_n/n=frac{1}{n}sum_{k=1}^n a_k$.
Then
$$x_n/n-c=left(frac{1}{n}sum_{k=1}^n a_kright) - c = frac{1}{n} sum_{k=1}^n (a_k-c)quad color{red} { (2) } $$
Let $epsilon>0$ and choose $N$ large enough such that $|a_n-c|<epsilon$ for all $nge N$. Then
$$left|frac{x_n}{n}-cright|le frac{1}{n}sum_{k=1}^N |a_k-c| + frac{n-N}{n}epsilon$$
for $nge N$. Now let $nrightarrowinfty$ while keeping $N,epsilon$ fixed. Then we obtain
$$limsup_{nrightarrowinfty}left|frac{x_n}{n}-cright|le epsilon$$
Since $epsilon$ was arbitrary, the conclusion follows.
Note:
In essence we just proved that a summable series is also Cesàro summable.
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
$endgroup$
We claim that $x_n/nrightarrow c$.
As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=sum_{k=1}^n a_k$ $color{red} { (1) }$. The assumption becomes $a_nrightarrow c$ and $x_n/n=frac{1}{n}sum_{k=1}^n a_k$.
Then
$$x_n/n-c=left(frac{1}{n}sum_{k=1}^n a_kright) - c = frac{1}{n} sum_{k=1}^n (a_k-c)quad color{red} { (2) } $$
Let $epsilon>0$ and choose $N$ large enough such that $|a_n-c|<epsilon$ for all $nge N$. Then
$$left|frac{x_n}{n}-cright|le frac{1}{n}sum_{k=1}^N |a_k-c| + frac{n-N}{n}epsilon$$
for $nge N$. Now let $nrightarrowinfty$ while keeping $N,epsilon$ fixed. Then we obtain
$$limsup_{nrightarrowinfty}left|frac{x_n}{n}-cright|le epsilon$$
Since $epsilon$ was arbitrary, the conclusion follows.
Note:
In essence we just proved that a summable series is also Cesàro summable.
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
edited May 31 '14 at 11:08
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
answered Feb 9 '14 at 19:18
J.R.J.R.
15.6k13055
15.6k13055
add a comment |
add a comment |
$begingroup$
That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$
$$lim_{n to infty} x_{n+1}-x_n = lim_{n to infty} frac{x_n}{n}$$
when $lim_{n to infty} x_{n+1}-x_n$ exists.
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
$endgroup$
add a comment |
$begingroup$
That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$
$$lim_{n to infty} x_{n+1}-x_n = lim_{n to infty} frac{x_n}{n}$$
when $lim_{n to infty} x_{n+1}-x_n$ exists.
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
$endgroup$
add a comment |
$begingroup$
That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$
$$lim_{n to infty} x_{n+1}-x_n = lim_{n to infty} frac{x_n}{n}$$
when $lim_{n to infty} x_{n+1}-x_n$ exists.
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
$endgroup$
That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$
$$lim_{n to infty} x_{n+1}-x_n = lim_{n to infty} frac{x_n}{n}$$
when $lim_{n to infty} x_{n+1}-x_n$ exists.
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
answered May 12 '14 at 16:53
Ehsan M. KermaniEhsan M. Kermani
6,36412348
6,36412348
add a comment |
add a comment |
$begingroup$
Apologies for editing Your Ad Here, but can someone please check (1) and (2)?
(1) = $sum_{1 le k le n} x_k - x_{k - 1}$ which telecopes to $-x_0 + x_n$, not just $x_n$ ?
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
$endgroup$
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
add a comment |
$begingroup$
Apologies for editing Your Ad Here, but can someone please check (1) and (2)?
(1) = $sum_{1 le k le n} x_k - x_{k - 1}$ which telecopes to $-x_0 + x_n$, not just $x_n$ ?
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
$endgroup$
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
add a comment |
$begingroup$
Apologies for editing Your Ad Here, but can someone please check (1) and (2)?
(1) = $sum_{1 le k le n} x_k - x_{k - 1}$ which telecopes to $-x_0 + x_n$, not just $x_n$ ?
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
$endgroup$
Apologies for editing Your Ad Here, but can someone please check (1) and (2)?
(1) = $sum_{1 le k le n} x_k - x_{k - 1}$ which telecopes to $-x_0 + x_n$, not just $x_n$ ?
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
answered May 12 '14 at 16:24
François MuerFrançois Muer
3561242
3561242
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
add a comment |
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
That is correct. I tacitly assumed we're talking about a sequence $x_1,x_2,dots$ and set $x_0=0$ for convenience. However, please note that your answer should rather be a comment on the previous answer.
$endgroup$
– J.R.
May 26 '14 at 20:05
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
$begingroup$
@YourAdHere thanks
$endgroup$
– François Muer
May 29 '14 at 20:31
add a comment |
$begingroup$
We let $a_n:=x_{n+1}-x_n$ and use this theorem to conclude that $$frac{a_1+a_2+...+a_n}{n}to c implies frac{(x_2-x_1)+(x_3-x_2)+....+(x_{n+1}-x_n)}{n}to c $$ $$ implies frac{x_{n+1}-x_1}{n}to c implies frac{x_n}{n}to c$$ since $x_1$ is finite.
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
$endgroup$
add a comment |
$begingroup$
We let $a_n:=x_{n+1}-x_n$ and use this theorem to conclude that $$frac{a_1+a_2+...+a_n}{n}to c implies frac{(x_2-x_1)+(x_3-x_2)+....+(x_{n+1}-x_n)}{n}to c $$ $$ implies frac{x_{n+1}-x_1}{n}to c implies frac{x_n}{n}to c$$ since $x_1$ is finite.
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
$endgroup$
add a comment |
$begingroup$
We let $a_n:=x_{n+1}-x_n$ and use this theorem to conclude that $$frac{a_1+a_2+...+a_n}{n}to c implies frac{(x_2-x_1)+(x_3-x_2)+....+(x_{n+1}-x_n)}{n}to c $$ $$ implies frac{x_{n+1}-x_1}{n}to c implies frac{x_n}{n}to c$$ since $x_1$ is finite.
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
$endgroup$
We let $a_n:=x_{n+1}-x_n$ and use this theorem to conclude that $$frac{a_1+a_2+...+a_n}{n}to c implies frac{(x_2-x_1)+(x_3-x_2)+....+(x_{n+1}-x_n)}{n}to c $$ $$ implies frac{x_{n+1}-x_1}{n}to c implies frac{x_n}{n}to c$$ since $x_1$ is finite.
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
answered Jan 7 at 9:18
RhaldrynRhaldryn
319414
319414
add a comment |
add a comment |
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3
$begingroup$
Rewrite it with $a_n:=x_n-x_{n-1}$ as $x_n=sum_{k=1}^n a_k$. So, the assumption is that $a_nto c$ and we are looking for the limit of the sequence of averages of $a_k$'s.
$endgroup$
– Berci
Feb 9 '14 at 18:27
2
$begingroup$
This is a special case of Stolz-Cesaro theorem. See Wikipedia, imomath or some posts on this site, for example this one.
$endgroup$
– Martin Sleziak
May 31 '14 at 12:01