Mixing
Divergent and convergent series of positive terms
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I recently read an article which contains the following facts :---
Let ${a_n}$ be a sequence of positive number such that $sum_{n=1}^{infty} a_n$ diverges, so we must have $a_nsim frac{1}{n^p}$ where $pleq 1$. Here $sim$ means nearby, for instance $frac{k}{n^p}$ for some constant $k$.
Another fact is that , let $sum_{n=1}^{infty} b_n$ be a series of positive terms and $sum_{n=1}^{infty} b_n$ is convergent, that means $a_n=O(frac{1}{n^p})$ for some $p>1$, where $O$ means order.
I cannot understand how do I prove these facts rigoursly, even I don't know whether they are true or not.
real-analysis sequences-and-series analysis divergent-series
asked Jan 7 at 18:06
MathloverMathlover
1426
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add a comment |
$begingroup$
I recently read an article which contains the following facts :---
Let ${a_n}$ be a sequence of positive number such that $sum_{n=1}^{infty} a_n$ diverges, so we must have $a_nsim frac{1}{n^p}$ where $pleq 1$. Here $sim$ means nearby, for instance $frac{k}{n^p}$ for some constant $k$.
Another fact is that , let $sum_{n=1}^{infty} b_n$ be a series of positive terms and $sum_{n=1}^{infty} b_n$ is convergent, that means $a_n=O(frac{1}{n^p})$ for some $p>1$, where $O$ means order.
I cannot understand how do I prove these facts rigoursly, even I don't know whether they are true or not.
real-analysis sequences-and-series analysis divergent-series
asked Jan 7 at 18:06
MathloverMathlover
1426
$endgroup$
add a comment |
$begingroup$
I recently read an article which contains the following facts :---
Let ${a_n}$ be a sequence of positive number such that $sum_{n=1}^{infty} a_n$ diverges, so we must have $a_nsim frac{1}{n^p}$ where $pleq 1$. Here $sim$ means nearby, for instance $frac{k}{n^p}$ for some constant $k$.
Another fact is that , let $sum_{n=1}^{infty} b_n$ be a series of positive terms and $sum_{n=1}^{infty} b_n$ is convergent, that means $a_n=O(frac{1}{n^p})$ for some $p>1$, where $O$ means order.
I cannot understand how do I prove these facts rigoursly, even I don't know whether they are true or not.
real-analysis sequences-and-series analysis divergent-series
asked Jan 7 at 18:06
MathloverMathlover
1426
$endgroup$
I recently read an article which contains the following facts :---
Let ${a_n}$ be a sequence of positive number such that $sum_{n=1}^{infty} a_n$ diverges, so we must have $a_nsim frac{1}{n^p}$ where $pleq 1$. Here $sim$ means nearby, for instance $frac{k}{n^p}$ for some constant $k$.
Another fact is that , let $sum_{n=1}^{infty} b_n$ be a series of positive terms and $sum_{n=1}^{infty} b_n$ is convergent, that means $a_n=O(frac{1}{n^p})$ for some $p>1$, where $O$ means order.
I cannot understand how do I prove these facts rigoursly, even I don't know whether they are true or not.
real-analysis sequences-and-series analysis divergent-series
real-analysis sequences-and-series analysis divergent-series
asked Jan 7 at 18:06
MathloverMathlover
1426
asked Jan 7 at 18:06
MathloverMathlover
1426
asked Jan 7 at 18:06
MathloverMathlover
1426
asked Jan 7 at 18:06
MathloverMathlover
1426
asked Jan 7 at 18:06
MathloverMathlover
1426
1426
add a comment |
add a comment |
1 Answer
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Either you are missing some hypotheses and/or context, or the article is blatantly wrong.
Take for instance $a_n=2^{-n}$ if $nneq 2^{p^2}$ for any integer $p$ and $a_{2^{p^2}}=frac{1}{p}$. The sum of $a_n$ is divergent, but there is no real number $a>0$ such that $a_n=O(n^{-a})$ or $n^{-a} = O(a_n)$ (however $a_n$ does go to $0$).
Take now the same sequence, except that $a_{2^{p^2}}=p^{-2}$. Then there is no real number $a>0$ such that $a_n=O(n^{-a})$ but the sum of the $a_n$ converges.
answered Jan 7 at 18:14
MindlackMindlack
3,36217
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Either you are missing some hypotheses and/or context, or the article is blatantly wrong.
Take for instance $a_n=2^{-n}$ if $nneq 2^{p^2}$ for any integer $p$ and $a_{2^{p^2}}=frac{1}{p}$. The sum of $a_n$ is divergent, but there is no real number $a>0$ such that $a_n=O(n^{-a})$ or $n^{-a} = O(a_n)$ (however $a_n$ does go to $0$).
Take now the same sequence, except that $a_{2^{p^2}}=p^{-2}$. Then there is no real number $a>0$ such that $a_n=O(n^{-a})$ but the sum of the $a_n$ converges.
answered Jan 7 at 18:14
MindlackMindlack
3,36217
$endgroup$
add a comment |
$begingroup$
Either you are missing some hypotheses and/or context, or the article is blatantly wrong.
Take for instance $a_n=2^{-n}$ if $nneq 2^{p^2}$ for any integer $p$ and $a_{2^{p^2}}=frac{1}{p}$. The sum of $a_n$ is divergent, but there is no real number $a>0$ such that $a_n=O(n^{-a})$ or $n^{-a} = O(a_n)$ (however $a_n$ does go to $0$).
Take now the same sequence, except that $a_{2^{p^2}}=p^{-2}$. Then there is no real number $a>0$ such that $a_n=O(n^{-a})$ but the sum of the $a_n$ converges.
answered Jan 7 at 18:14
MindlackMindlack
3,36217
$endgroup$
add a comment |
$begingroup$
Either you are missing some hypotheses and/or context, or the article is blatantly wrong.
Take for instance $a_n=2^{-n}$ if $nneq 2^{p^2}$ for any integer $p$ and $a_{2^{p^2}}=frac{1}{p}$. The sum of $a_n$ is divergent, but there is no real number $a>0$ such that $a_n=O(n^{-a})$ or $n^{-a} = O(a_n)$ (however $a_n$ does go to $0$).
Take now the same sequence, except that $a_{2^{p^2}}=p^{-2}$. Then there is no real number $a>0$ such that $a_n=O(n^{-a})$ but the sum of the $a_n$ converges.
answered Jan 7 at 18:14
MindlackMindlack
3,36217
$endgroup$
Either you are missing some hypotheses and/or context, or the article is blatantly wrong.
Take for instance $a_n=2^{-n}$ if $nneq 2^{p^2}$ for any integer $p$ and $a_{2^{p^2}}=frac{1}{p}$. The sum of $a_n$ is divergent, but there is no real number $a>0$ such that $a_n=O(n^{-a})$ or $n^{-a} = O(a_n)$ (however $a_n$ does go to $0$).
Take now the same sequence, except that $a_{2^{p^2}}=p^{-2}$. Then there is no real number $a>0$ such that $a_n=O(n^{-a})$ but the sum of the $a_n$ converges.
answered Jan 7 at 18:14
MindlackMindlack
3,36217
answered Jan 7 at 18:14
MindlackMindlack
3,36217
answered Jan 7 at 18:14
MindlackMindlack
3,36217
answered Jan 7 at 18:14
MindlackMindlack
3,36217
3,36217
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