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Infinite sum of harmonic number


Sum of reciprocals of numbers with certain terms omittedWhat is the closed form sum of this series?Is there a name for infinite series of this type?Infinite amount of additions, finite sum?Why can infinite series be summed different ways to get different results?Why does the order of summation of the terms of an infinite series influence its value?Infinite Series $left(frac12+frac14-frac23right)+left(frac15+frac17-frac26right)+left(frac18+frac110-frac29right)+cdots$Sum of the recripocals of the Harmonic NumbersFind the sum of the infinite seriesAlternating harmonic series convergence













4












$begingroup$


I learned that I can find the value of some infinite sum.



Then what is the value of this sum?
$$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
And I want to know How to find the value of the infinite sum of this-like form.






sequences-and-series







share|cite|improve this question









$endgroup$
















    4












    $begingroup$


    I learned that I can find the value of some infinite sum.



    Then what is the value of this sum?
    $$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
    And I want to know How to find the value of the infinite sum of this-like form.






    sequences-and-series







    share|cite|improve this question









    $endgroup$














      4












      4








      4


      3



      $begingroup$


      I learned that I can find the value of some infinite sum.



      Then what is the value of this sum?
      $$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
      And I want to know How to find the value of the infinite sum of this-like form.






      sequences-and-series







      share|cite|improve this question









      $endgroup$




      I learned that I can find the value of some infinite sum.



      Then what is the value of this sum?
      $$frac12 + left(1+frac12right)frac12^2+left(1+frac12 +frac13right)frac12^3+left(1+frac12 +frac13 +frac14right)frac12^4 + cdots $$
      And I want to know How to find the value of the infinite sum of this-like form.






      sequences-and-series




      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      S. YooS. Yoo

      424




      424




















          1 Answer
          1






          active

          oldest

          votes


















          8












          $begingroup$

          One can split this summation into
          $$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
          $$beginalign
          &=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
          &=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
          &=sum_k=1^infty frac2^1-kk\
          &=2sum_k=1^infty fracleft(frac12right)^kk\
          &=2left(-lnleft(1-frac12right)right)\
          &boxed=2ln(2)
          endalign$$
          By using the fact that
          $$ln(1-x)=-sum_k=1^infty fracx^kk$$
          for all $|x|lt 1$.



          In fact one can use a similar method to prove that
          $$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
          for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
          $$H_k=sum_n=1^kfrac1n$$






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            active

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            8












            $begingroup$

            One can split this summation into
            $$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
            $$beginalign
            &=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
            &=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
            &=sum_k=1^infty frac2^1-kk\
            &=2sum_k=1^infty fracleft(frac12right)^kk\
            &=2left(-lnleft(1-frac12right)right)\
            &boxed=2ln(2)
            endalign$$
            By using the fact that
            $$ln(1-x)=-sum_k=1^infty fracx^kk$$
            for all $|x|lt 1$.



            In fact one can use a similar method to prove that
            $$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
            for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
            $$H_k=sum_n=1^kfrac1n$$






            share|cite|improve this answer











            $endgroup$

















              8












              $begingroup$

              One can split this summation into
              $$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
              $$beginalign
              &=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
              &=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
              &=sum_k=1^infty frac2^1-kk\
              &=2sum_k=1^infty fracleft(frac12right)^kk\
              &=2left(-lnleft(1-frac12right)right)\
              &boxed=2ln(2)
              endalign$$
              By using the fact that
              $$ln(1-x)=-sum_k=1^infty fracx^kk$$
              for all $|x|lt 1$.



              In fact one can use a similar method to prove that
              $$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
              for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
              $$H_k=sum_n=1^kfrac1n$$






              share|cite|improve this answer











              $endgroup$















                8












                8








                8





                $begingroup$

                One can split this summation into
                $$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
                $$beginalign
                &=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
                &=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
                &=sum_k=1^infty frac2^1-kk\
                &=2sum_k=1^infty fracleft(frac12right)^kk\
                &=2left(-lnleft(1-frac12right)right)\
                &boxed=2ln(2)
                endalign$$
                By using the fact that
                $$ln(1-x)=-sum_k=1^infty fracx^kk$$
                for all $|x|lt 1$.



                In fact one can use a similar method to prove that
                $$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
                for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
                $$H_k=sum_n=1^kfrac1n$$






                share|cite|improve this answer











                $endgroup$



                One can split this summation into
                $$left(frac12+frac12^2+frac12^3+cdotsright)+frac12left(frac12^2+frac12^3+cdotsright)+frac13left(frac12^3+frac12^4+cdotsright)+cdots$$
                $$beginalign
                &=sum_k=1^infty frac1k sum_j=k^infty frac12^j\
                &=sum_k=1^infty frac1k left(frac2^-k1-2^-1right)\
                &=sum_k=1^infty frac2^1-kk\
                &=2sum_k=1^infty fracleft(frac12right)^kk\
                &=2left(-lnleft(1-frac12right)right)\
                &boxed=2ln(2)
                endalign$$
                By using the fact that
                $$ln(1-x)=-sum_k=1^infty fracx^kk$$
                for all $|x|lt 1$.



                In fact one can use a similar method to prove that
                $$sum_k=1^infty x^kH_k=frac11-xlnleft(frac11-xright)$$
                for $|x|lt1$. Where $H_k$ is the $k$th harmonic number given by
                $$H_k=sum_n=1^kfrac1n$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 3 hours ago









                Théophile

                20.3k13047




                20.3k13047










                answered 7 hours ago









                Peter ForemanPeter Foreman

                5,5291216




                5,5291216



























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