Prove that for $a_k>0,$ if $sum a_k^2$ converges, then $sum frac{a_k}k$ converges. [duplicate]












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  • Converging series question, Prove that if $sum_{n=1}^{infty} a_n^{2}$ converges, then does $sum_{n=1}^{infty} frac {a_n}{n}$

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Prove that for for $a_k>0,$ if $sum a_k^2$ converges, then $sum frac{a_k}k$ converges.
I was given this in an introductory calculus class, where I was only taught the basic convergence tests. I’ve tried limit comparison, power series, direct comparison, all to no avail. I have tried proving the contrapositive, using integrals as well, but the limit comparison with any series I’ve tried just goes to 0 or infinity which is inconclusive. My searches on MSE just yield the simpler problem of “if$sum a_k$ converges then prove $sum a_k^2$ converges“, and searches on google turned up nothing. Thank you for any help.






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 Jan 31 at 22:37


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    I suppose you are not allowed to use Cauchy-Schwartz inequality en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality ?
    $endgroup$
    – Tito Eliatron
    Jan 31 at 22:28






  • 1




    $begingroup$
    No need here for the full Cauchy-Schwarz, just note that $frac{a_k}{k} leq a_k^2+k^{-2}$.
    $endgroup$
    – Mindlack
    Jan 31 at 22:32
















1












$begingroup$



This question already has an answer here:




  • Converging series question, Prove that if $sum_{n=1}^{infty} a_n^{2}$ converges, then does $sum_{n=1}^{infty} frac {a_n}{n}$

    3 answers




Prove that for for $a_k>0,$ if $sum a_k^2$ converges, then $sum frac{a_k}k$ converges.
I was given this in an introductory calculus class, where I was only taught the basic convergence tests. I’ve tried limit comparison, power series, direct comparison, all to no avail. I have tried proving the contrapositive, using integrals as well, but the limit comparison with any series I’ve tried just goes to 0 or infinity which is inconclusive. My searches on MSE just yield the simpler problem of “if$sum a_k$ converges then prove $sum a_k^2$ converges“, and searches on google turned up nothing. Thank you for any help.






sequences-and-series







share|cite|improve this question











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marked as duplicate by Robert Wolfe, Mark Viola sequences-and-series
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 Jan 31 at 22:37


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    I suppose you are not allowed to use Cauchy-Schwartz inequality en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality ?
    $endgroup$
    – Tito Eliatron
    Jan 31 at 22:28






  • 1




    $begingroup$
    No need here for the full Cauchy-Schwarz, just note that $frac{a_k}{k} leq a_k^2+k^{-2}$.
    $endgroup$
    – Mindlack
    Jan 31 at 22:32














1












1








1


1



$begingroup$



This question already has an answer here:




  • Converging series question, Prove that if $sum_{n=1}^{infty} a_n^{2}$ converges, then does $sum_{n=1}^{infty} frac {a_n}{n}$

    3 answers




Prove that for for $a_k>0,$ if $sum a_k^2$ converges, then $sum frac{a_k}k$ converges.
I was given this in an introductory calculus class, where I was only taught the basic convergence tests. I’ve tried limit comparison, power series, direct comparison, all to no avail. I have tried proving the contrapositive, using integrals as well, but the limit comparison with any series I’ve tried just goes to 0 or infinity which is inconclusive. My searches on MSE just yield the simpler problem of “if$sum a_k$ converges then prove $sum a_k^2$ converges“, and searches on google turned up nothing. Thank you for any help.






sequences-and-series







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Converging series question, Prove that if $sum_{n=1}^{infty} a_n^{2}$ converges, then does $sum_{n=1}^{infty} frac {a_n}{n}$

    3 answers




Prove that for for $a_k>0,$ if $sum a_k^2$ converges, then $sum frac{a_k}k$ converges.
I was given this in an introductory calculus class, where I was only taught the basic convergence tests. I’ve tried limit comparison, power series, direct comparison, all to no avail. I have tried proving the contrapositive, using integrals as well, but the limit comparison with any series I’ve tried just goes to 0 or infinity which is inconclusive. My searches on MSE just yield the simpler problem of “if$sum a_k$ converges then prove $sum a_k^2$ converges“, and searches on google turned up nothing. Thank you for any help.





This question already has an answer here:




  • Converging series question, Prove that if $sum_{n=1}^{infty} a_n^{2}$ converges, then does $sum_{n=1}^{infty} frac {a_n}{n}$

    3 answers







sequences-and-series




sequences-and-series






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edited Jan 31 at 22:42









Bernard

124k741117




124k741117










asked Jan 31 at 22:25









D.R.D.R.

1,781823




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marked as duplicate by Robert Wolfe, Mark Viola sequences-and-series
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 Jan 31 at 22:37


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Robert Wolfe, Mark Viola sequences-and-series
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 Jan 31 at 22:37


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    I suppose you are not allowed to use Cauchy-Schwartz inequality en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality ?
    $endgroup$
    – Tito Eliatron
    Jan 31 at 22:28






  • 1




    $begingroup$
    No need here for the full Cauchy-Schwarz, just note that $frac{a_k}{k} leq a_k^2+k^{-2}$.
    $endgroup$
    – Mindlack
    Jan 31 at 22:32














  • 1




    $begingroup$
    I suppose you are not allowed to use Cauchy-Schwartz inequality en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality ?
    $endgroup$
    – Tito Eliatron
    Jan 31 at 22:28






  • 1




    $begingroup$
    No need here for the full Cauchy-Schwarz, just note that $frac{a_k}{k} leq a_k^2+k^{-2}$.
    $endgroup$
    – Mindlack
    Jan 31 at 22:32








1




1




$begingroup$
I suppose you are not allowed to use Cauchy-Schwartz inequality en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality ?
$endgroup$
– Tito Eliatron
Jan 31 at 22:28




$begingroup$
I suppose you are not allowed to use Cauchy-Schwartz inequality en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality ?
$endgroup$
– Tito Eliatron
Jan 31 at 22:28




1




1




$begingroup$
No need here for the full Cauchy-Schwarz, just note that $frac{a_k}{k} leq a_k^2+k^{-2}$.
$endgroup$
– Mindlack
Jan 31 at 22:32




$begingroup$
No need here for the full Cauchy-Schwarz, just note that $frac{a_k}{k} leq a_k^2+k^{-2}$.
$endgroup$
– Mindlack
Jan 31 at 22:32










1 Answer
1






active

oldest

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$begingroup$

$a^2+b^2-2ab=(a-b)^2ge0$ hence, $able frac{a^2+b^2}{2}$. Hence, for any $n$ you have that $0le frac{a_n}{n}lefrac{a_n^2+frac{1}{n^2}}{2}$, so
$$0le sum_{n=1}^infty frac{a_n}{n}lefrac{1}{2}left(sum_{n=1}^infty a_n^2+sum_{n=1}^infty frac{1}{n^2}right).$$
So, if $sum a_n^2$ converges, so it does $sumfrac{a_n}{n}$.






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$endgroup$









  • 1




    $begingroup$
    Since $sum n^{-2}=pi^2/6$ converges.
    $endgroup$
    – Antinous
    Jan 31 at 22:39


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

$a^2+b^2-2ab=(a-b)^2ge0$ hence, $able frac{a^2+b^2}{2}$. Hence, for any $n$ you have that $0le frac{a_n}{n}lefrac{a_n^2+frac{1}{n^2}}{2}$, so
$$0le sum_{n=1}^infty frac{a_n}{n}lefrac{1}{2}left(sum_{n=1}^infty a_n^2+sum_{n=1}^infty frac{1}{n^2}right).$$
So, if $sum a_n^2$ converges, so it does $sumfrac{a_n}{n}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Since $sum n^{-2}=pi^2/6$ converges.
    $endgroup$
    – Antinous
    Jan 31 at 22:39
















6












$begingroup$

$a^2+b^2-2ab=(a-b)^2ge0$ hence, $able frac{a^2+b^2}{2}$. Hence, for any $n$ you have that $0le frac{a_n}{n}lefrac{a_n^2+frac{1}{n^2}}{2}$, so
$$0le sum_{n=1}^infty frac{a_n}{n}lefrac{1}{2}left(sum_{n=1}^infty a_n^2+sum_{n=1}^infty frac{1}{n^2}right).$$
So, if $sum a_n^2$ converges, so it does $sumfrac{a_n}{n}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Since $sum n^{-2}=pi^2/6$ converges.
    $endgroup$
    – Antinous
    Jan 31 at 22:39














6












6








6





$begingroup$

$a^2+b^2-2ab=(a-b)^2ge0$ hence, $able frac{a^2+b^2}{2}$. Hence, for any $n$ you have that $0le frac{a_n}{n}lefrac{a_n^2+frac{1}{n^2}}{2}$, so
$$0le sum_{n=1}^infty frac{a_n}{n}lefrac{1}{2}left(sum_{n=1}^infty a_n^2+sum_{n=1}^infty frac{1}{n^2}right).$$
So, if $sum a_n^2$ converges, so it does $sumfrac{a_n}{n}$.






share|cite|improve this answer









$endgroup$



$a^2+b^2-2ab=(a-b)^2ge0$ hence, $able frac{a^2+b^2}{2}$. Hence, for any $n$ you have that $0le frac{a_n}{n}lefrac{a_n^2+frac{1}{n^2}}{2}$, so
$$0le sum_{n=1}^infty frac{a_n}{n}lefrac{1}{2}left(sum_{n=1}^infty a_n^2+sum_{n=1}^infty frac{1}{n^2}right).$$
So, if $sum a_n^2$ converges, so it does $sumfrac{a_n}{n}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 31 at 22:33









Tito EliatronTito Eliatron

1,609622




1,609622








  • 1




    $begingroup$
    Since $sum n^{-2}=pi^2/6$ converges.
    $endgroup$
    – Antinous
    Jan 31 at 22:39














  • 1




    $begingroup$
    Since $sum n^{-2}=pi^2/6$ converges.
    $endgroup$
    – Antinous
    Jan 31 at 22:39








1




1




$begingroup$
Since $sum n^{-2}=pi^2/6$ converges.
$endgroup$
– Antinous
Jan 31 at 22:39




$begingroup$
Since $sum n^{-2}=pi^2/6$ converges.
$endgroup$
– Antinous
Jan 31 at 22:39



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