Convergence of a series involving logs $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$












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I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.






sequences-and-series







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  • 4




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    You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
    $endgroup$
    – p4sch
    Jan 14 at 20:27










  • $begingroup$
    Could you write down how did you reduce the series from first form to the second one?
    $endgroup$
    – user
    Jan 14 at 22:07










  • $begingroup$
    In the denominator, is it the log squared or the log of the square?
    $endgroup$
    – Bernard
    Jan 14 at 22:28
















0












$begingroup$


I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
    $endgroup$
    – p4sch
    Jan 14 at 20:27










  • $begingroup$
    Could you write down how did you reduce the series from first form to the second one?
    $endgroup$
    – user
    Jan 14 at 22:07










  • $begingroup$
    In the denominator, is it the log squared or the log of the square?
    $endgroup$
    – Bernard
    Jan 14 at 22:28














0












0








0





$begingroup$


I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$




I am working on showing the convergence of the series $sum_{n=2}^{infty} frac{sin(nx)}{ln(n)}$ and have reduced the problem to showing that $sum_{n=2}^{infty} frac{ln(frac{n}{n+1})}{ln(n)^2}$ converges. I have a hunch that it does converge, but am having trouble manipulating it to something I can apply the comparison test to. Hints to get me started are appreciated.






sequences-and-series




sequences-and-series






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share|cite|improve this question













share|cite|improve this question




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edited Jan 14 at 21:58









Bernard

121k740116




121k740116










asked Jan 14 at 20:25









JungleshrimpJungleshrimp

314111




314111








  • 4




    $begingroup$
    You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
    $endgroup$
    – p4sch
    Jan 14 at 20:27










  • $begingroup$
    Could you write down how did you reduce the series from first form to the second one?
    $endgroup$
    – user
    Jan 14 at 22:07










  • $begingroup$
    In the denominator, is it the log squared or the log of the square?
    $endgroup$
    – Bernard
    Jan 14 at 22:28














  • 4




    $begingroup$
    You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
    $endgroup$
    – p4sch
    Jan 14 at 20:27










  • $begingroup$
    Could you write down how did you reduce the series from first form to the second one?
    $endgroup$
    – user
    Jan 14 at 22:07










  • $begingroup$
    In the denominator, is it the log squared or the log of the square?
    $endgroup$
    – Bernard
    Jan 14 at 22:28








4




4




$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27




$begingroup$
You should use that $- ln(n/(n+1)) =ln(1+1/n) sim 1/n$.
$endgroup$
– p4sch
Jan 14 at 20:27












$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07




$begingroup$
Could you write down how did you reduce the series from first form to the second one?
$endgroup$
– user
Jan 14 at 22:07












$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28




$begingroup$
In the denominator, is it the log squared or the log of the square?
$endgroup$
– Bernard
Jan 14 at 22:28










1 Answer
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Supposing the denominator is the log squared, we can use equivalence:
$$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
which is Bertrand's series.



This Bertrand's series converges by the integral test.






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    1 Answer
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    $begingroup$

    Supposing the denominator is the log squared, we can use equivalence:
    $$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
    which is Bertrand's series.



    This Bertrand's series converges by the integral test.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Supposing the denominator is the log squared, we can use equivalence:
      $$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
      which is Bertrand's series.



      This Bertrand's series converges by the integral test.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Supposing the denominator is the log squared, we can use equivalence:
        $$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
        which is Bertrand's series.



        This Bertrand's series converges by the integral test.






        share|cite|improve this answer









        $endgroup$



        Supposing the denominator is the log squared, we can use equivalence:
        $$frac{ln(dfrac{n}{n+1})}{ln^2 n}=-frac{lnBigl(1+dfrac1{n}Bigr)}{ln^2 n}sim_infty-frac{dfrac1{n}}{ln^2n}=-frac{1}{n,ln^2 n},$$
        which is Bertrand's series.



        This Bertrand's series converges by the integral test.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 22:46









        BernardBernard

        121k740116




        121k740116






























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