Problem related to convergence of a series












1












$begingroup$


Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series







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$endgroup$








  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04
















1












$begingroup$


Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04














1












1








1





$begingroup$


Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series







share|cite|improve this question











$endgroup$




Problem




Determine whether the given series is convergent or not:
$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)$$




Attempt



Checking for absolute convergence



$|sum_{1}^{infty}log(nsin(frac{1}{n}))|=|sum_{1}^{infty}(log(n(frac{1}{n}-frac{-1}{n^3 3!}+...)))|
=|sum_{1}^{infty}(frac{-1}{n^2 3!}+frac{-1}{n^4 5!}+...)|
=|sum_{1}^{infty}(frac{1}{n^2 3!}-frac{-1}{n^4 5!}+...)|$



Comparing with $sum_{1}^{infty}frac{1}{n^2}$ , the series converges absolutely




Is this correct?







calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 6:14









Bungo

13.6k22148




13.6k22148










asked Dec 17 '18 at 5:13









blue boyblue boy

1,236613




1,236613








  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04














  • 1




    $begingroup$
    You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 6:02










  • $begingroup$
    Oh my bad . Sorry.
    $endgroup$
    – blue boy
    Dec 17 '18 at 6:04








1




1




$begingroup$
You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 6:02




$begingroup$
You ask if 'function is convergent or not', then talk about some series. Please clarify whether you are interested in a sequence or a series.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 6:02












$begingroup$
Oh my bad . Sorry.
$endgroup$
– blue boy
Dec 17 '18 at 6:04




$begingroup$
Oh my bad . Sorry.
$endgroup$
– blue boy
Dec 17 '18 at 6:04










2 Answers
2






active

oldest

votes


















1












$begingroup$

The idea is correct but we don't need to use absolute convergence, we have indeed that



$$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



and both series converge.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Errors:





    1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





      1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




    It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



    It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You had the right idea but the devil is in the details.
      $endgroup$
      – DanielWainfleet
      Dec 17 '18 at 6:14










    • $begingroup$
      I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
      $endgroup$
      – blue boy
      Dec 17 '18 at 6:15











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The idea is correct but we don't need to use absolute convergence, we have indeed that



    $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



    and both series converge.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The idea is correct but we don't need to use absolute convergence, we have indeed that



      $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



      and both series converge.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The idea is correct but we don't need to use absolute convergence, we have indeed that



        $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



        and both series converge.






        share|cite|improve this answer









        $endgroup$



        The idea is correct but we don't need to use absolute convergence, we have indeed that



        $$sum_{n=1}^{infty}logleft(nsinleft(frac{1}{n}right)right)=sum_{n=1}^{infty}logleft(1-frac1{6n^2}+Oleft(frac1{n^4}right)right)=-frac16sum_{n=1}^{infty}frac1{n^2}+sum_{n=1}^{infty}Oleft(frac1{n^4}right)$$



        and both series converge.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 7:34









        gimusigimusi

        92.9k94494




        92.9k94494























            2












            $begingroup$

            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15
















            2












            $begingroup$

            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15














            2












            2








            2





            $begingroup$

            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.






            share|cite|improve this answer









            $endgroup$



            Errors:





            1. The second "$=$" sign is wrong. We do not have $|log (1+x)|=|x|$ unless $x=0.$





              1. $log (nsin 1/n)=log (1-d)=-(d+d^2/2+d^3/3+...)$ where $d=1/3!n^2-1/5!n^4+-...$




            It suffices to observe that $0<d<1/3!n^2<1$ when $nin Bbb N,$ as the terms in the series for $d$ are alternating and diminishing, so that $0<d+d^2/2+d^3/3+...<d+d^2+d^3+...=$ $=d/(1-d)<d/(1-1/3!n^2)leq d/(1-1/3!)<(1/3!n^2)/(1-1/3!)=O(1/n^2).$



            It would be better if you edited to say " the series $sum_{n=1}^{infty}log (nsin 1/n)$ " in the first line.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '18 at 6:11









            DanielWainfleetDanielWainfleet

            34.7k31648




            34.7k31648












            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15


















            • $begingroup$
              You had the right idea but the devil is in the details.
              $endgroup$
              – DanielWainfleet
              Dec 17 '18 at 6:14










            • $begingroup$
              I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
              $endgroup$
              – blue boy
              Dec 17 '18 at 6:15
















            $begingroup$
            You had the right idea but the devil is in the details.
            $endgroup$
            – DanielWainfleet
            Dec 17 '18 at 6:14




            $begingroup$
            You had the right idea but the devil is in the details.
            $endgroup$
            – DanielWainfleet
            Dec 17 '18 at 6:14












            $begingroup$
            I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
            $endgroup$
            – blue boy
            Dec 17 '18 at 6:15




            $begingroup$
            I think it is not clear. I am expanding $log(1-x)$ and not considering the higher order terms.
            $endgroup$
            – blue boy
            Dec 17 '18 at 6:15


















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