Prove convergence from $|sum_{k = 1}^n a_k |leq sqrt{n}$ [duplicate]











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  • Does $big|sum_{k=1}^na_kbig|leqsqrt{n}$ imply the convergence of $sum_{1}^infty frac{a_k}{k}$?

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Given $|sum_{k = 1}^n a_k |leq sqrt{n}$, how to show that $sum_{k = 1}^infty a_k/k $ converges?.



I've not encountered a convergence problem where I don't know the convergence of $(a_n)$. So the usual test are out of option.






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    Use Dirichlet's test
    – Yadati Kiran
    Nov 15 at 7:44










  • @YadatiKiran, but $M$ is a constant the bound here is changing for each $n$
    – Hawk
    Nov 15 at 7:45















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  • Does $big|sum_{k=1}^na_kbig|leqsqrt{n}$ imply the convergence of $sum_{1}^infty frac{a_k}{k}$?

    1 answer




Given $|sum_{k = 1}^n a_k |leq sqrt{n}$, how to show that $sum_{k = 1}^infty a_k/k $ converges?.



I've not encountered a convergence problem where I don't know the convergence of $(a_n)$. So the usual test are out of option.






real-analysis







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 Nov 15 at 7:54


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  • 1




    Use Dirichlet's test
    – Yadati Kiran
    Nov 15 at 7:44










  • @YadatiKiran, but $M$ is a constant the bound here is changing for each $n$
    – Hawk
    Nov 15 at 7:45













up vote
1
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1









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1






This question already has an answer here:




  • Does $big|sum_{k=1}^na_kbig|leqsqrt{n}$ imply the convergence of $sum_{1}^infty frac{a_k}{k}$?

    1 answer




Given $|sum_{k = 1}^n a_k |leq sqrt{n}$, how to show that $sum_{k = 1}^infty a_k/k $ converges?.



I've not encountered a convergence problem where I don't know the convergence of $(a_n)$. So the usual test are out of option.






real-analysis







share|cite|improve this question














This question already has an answer here:




  • Does $big|sum_{k=1}^na_kbig|leqsqrt{n}$ imply the convergence of $sum_{1}^infty frac{a_k}{k}$?

    1 answer




Given $|sum_{k = 1}^n a_k |leq sqrt{n}$, how to show that $sum_{k = 1}^infty a_k/k $ converges?.



I've not encountered a convergence problem where I don't know the convergence of $(a_n)$. So the usual test are out of option.





This question already has an answer here:




  • Does $big|sum_{k=1}^na_kbig|leqsqrt{n}$ imply the convergence of $sum_{1}^infty frac{a_k}{k}$?

    1 answer







real-analysis




real-analysis






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asked Nov 15 at 7:38









Hawk

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marked as duplicate by Robert Z real-analysis
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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




    Use Dirichlet's test
    – Yadati Kiran
    Nov 15 at 7:44










  • @YadatiKiran, but $M$ is a constant the bound here is changing for each $n$
    – Hawk
    Nov 15 at 7:45














  • 1




    Use Dirichlet's test
    – Yadati Kiran
    Nov 15 at 7:44










  • @YadatiKiran, but $M$ is a constant the bound here is changing for each $n$
    – Hawk
    Nov 15 at 7:45








1




1




Use Dirichlet's test
– Yadati Kiran
Nov 15 at 7:44




Use Dirichlet's test
– Yadati Kiran
Nov 15 at 7:44












@YadatiKiran, but $M$ is a constant the bound here is changing for each $n$
– Hawk
Nov 15 at 7:45




@YadatiKiran, but $M$ is a constant the bound here is changing for each $n$
– Hawk
Nov 15 at 7:45










1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










Let $displaystyle s_n=sum_{k=1}^{n} a_k$. $:$ Then $displaystylesum_{k=1}^{n} dfrac {a_k} k = sum_{k=1}^{n} dfrac {s_k-s_{k-1}} k $ (with the convention $s_0=0$). $:$ This can be rewritten as $displaystylesum_{k=1}^{n-1} s_k (frac 1 k -frac 1 {k+1})+frac {s_n} n$. $:$ It remains only to observe that $displaystylesum_{k=1}^{infty} sqrt k (frac 1 k -frac 1 {k+1})=sum_{k=1}^{infty} sqrt k frac 1 {k(k+1)}$ is convergent by comparison with $displaystylesum k^{-3/2}$.






share|cite|improve this answer























  • Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
    – Hawk
    Nov 15 at 7:48






  • 1




    @Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
    – Kavi Rama Murthy
    Nov 15 at 7:52












  • Thank you for showing me this trick.
    – Hawk
    Nov 15 at 7:53


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Let $displaystyle s_n=sum_{k=1}^{n} a_k$. $:$ Then $displaystylesum_{k=1}^{n} dfrac {a_k} k = sum_{k=1}^{n} dfrac {s_k-s_{k-1}} k $ (with the convention $s_0=0$). $:$ This can be rewritten as $displaystylesum_{k=1}^{n-1} s_k (frac 1 k -frac 1 {k+1})+frac {s_n} n$. $:$ It remains only to observe that $displaystylesum_{k=1}^{infty} sqrt k (frac 1 k -frac 1 {k+1})=sum_{k=1}^{infty} sqrt k frac 1 {k(k+1)}$ is convergent by comparison with $displaystylesum k^{-3/2}$.






share|cite|improve this answer























  • Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
    – Hawk
    Nov 15 at 7:48






  • 1




    @Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
    – Kavi Rama Murthy
    Nov 15 at 7:52












  • Thank you for showing me this trick.
    – Hawk
    Nov 15 at 7:53















up vote
5
down vote



accepted










Let $displaystyle s_n=sum_{k=1}^{n} a_k$. $:$ Then $displaystylesum_{k=1}^{n} dfrac {a_k} k = sum_{k=1}^{n} dfrac {s_k-s_{k-1}} k $ (with the convention $s_0=0$). $:$ This can be rewritten as $displaystylesum_{k=1}^{n-1} s_k (frac 1 k -frac 1 {k+1})+frac {s_n} n$. $:$ It remains only to observe that $displaystylesum_{k=1}^{infty} sqrt k (frac 1 k -frac 1 {k+1})=sum_{k=1}^{infty} sqrt k frac 1 {k(k+1)}$ is convergent by comparison with $displaystylesum k^{-3/2}$.






share|cite|improve this answer























  • Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
    – Hawk
    Nov 15 at 7:48






  • 1




    @Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
    – Kavi Rama Murthy
    Nov 15 at 7:52












  • Thank you for showing me this trick.
    – Hawk
    Nov 15 at 7:53













up vote
5
down vote



accepted







up vote
5
down vote



accepted






Let $displaystyle s_n=sum_{k=1}^{n} a_k$. $:$ Then $displaystylesum_{k=1}^{n} dfrac {a_k} k = sum_{k=1}^{n} dfrac {s_k-s_{k-1}} k $ (with the convention $s_0=0$). $:$ This can be rewritten as $displaystylesum_{k=1}^{n-1} s_k (frac 1 k -frac 1 {k+1})+frac {s_n} n$. $:$ It remains only to observe that $displaystylesum_{k=1}^{infty} sqrt k (frac 1 k -frac 1 {k+1})=sum_{k=1}^{infty} sqrt k frac 1 {k(k+1)}$ is convergent by comparison with $displaystylesum k^{-3/2}$.






share|cite|improve this answer














Let $displaystyle s_n=sum_{k=1}^{n} a_k$. $:$ Then $displaystylesum_{k=1}^{n} dfrac {a_k} k = sum_{k=1}^{n} dfrac {s_k-s_{k-1}} k $ (with the convention $s_0=0$). $:$ This can be rewritten as $displaystylesum_{k=1}^{n-1} s_k (frac 1 k -frac 1 {k+1})+frac {s_n} n$. $:$ It remains only to observe that $displaystylesum_{k=1}^{infty} sqrt k (frac 1 k -frac 1 {k+1})=sum_{k=1}^{infty} sqrt k frac 1 {k(k+1)}$ is convergent by comparison with $displaystylesum k^{-3/2}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 15 at 7:49









Yadati Kiran

954316




954316










answered Nov 15 at 7:45









Kavi Rama Murthy

41.4k31751




41.4k31751












  • Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
    – Hawk
    Nov 15 at 7:48






  • 1




    @Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
    – Kavi Rama Murthy
    Nov 15 at 7:52












  • Thank you for showing me this trick.
    – Hawk
    Nov 15 at 7:53


















  • Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
    – Hawk
    Nov 15 at 7:48






  • 1




    @Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
    – Kavi Rama Murthy
    Nov 15 at 7:52












  • Thank you for showing me this trick.
    – Hawk
    Nov 15 at 7:53
















Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
– Hawk
Nov 15 at 7:48




Is there an instance when we know to invoke this telescoping $sum_{k=1}^{n} frac {a_k} k = sum_{k=1}^{n} frac {s_k-s_{k-1}} k$ technique?
– Hawk
Nov 15 at 7:48




1




1




@Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
– Kavi Rama Murthy
Nov 15 at 7:52






@Hawk The fact that we are given abound on the sums $s_k$ and not directly on the coefficients suggests that you should write $a_k=s_k-s_{k-1}$ and proceed as I have done. This technique is very powerful (especially in Fourier series).
– Kavi Rama Murthy
Nov 15 at 7:52














Thank you for showing me this trick.
– Hawk
Nov 15 at 7:53




Thank you for showing me this trick.
– Hawk
Nov 15 at 7:53



-

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