Infinite Series $1+frac12-frac23+frac14+frac15-frac26+cdots$











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Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $ln(1+x)$ and $arctan(x)$:
$$1+frac12-frac23+frac14+frac15-frac26+cdots$$



The question isn't homework or anything, just a thought tease.
I tried for a long while but couldn't find anything remotely close.
Thanks in advance for the help.










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  • 3




    The partial sum of the first $3n$ terms are $H_{3n} - H_{n}$ where $H_n$ is the $n^{th}$ harmonic number, so the answer is $log 3$.
    – achille hui
    Jan 30 '14 at 13:58










  • A comment to sequence 1,1,-2,1,1,-2,... oeis.org/A061347 includes the generalization to log(n). See also math.stackexchange.com/questions/46378/… and math.stackexchange.com/questions/883348/series-for-logarithms
    – Jaume Oliver Lafont
    Jan 12 '16 at 22:20















up vote
6
down vote

favorite
6












Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $ln(1+x)$ and $arctan(x)$:
$$1+frac12-frac23+frac14+frac15-frac26+cdots$$



The question isn't homework or anything, just a thought tease.
I tried for a long while but couldn't find anything remotely close.
Thanks in advance for the help.










share|cite|improve this question




















  • 3




    The partial sum of the first $3n$ terms are $H_{3n} - H_{n}$ where $H_n$ is the $n^{th}$ harmonic number, so the answer is $log 3$.
    – achille hui
    Jan 30 '14 at 13:58










  • A comment to sequence 1,1,-2,1,1,-2,... oeis.org/A061347 includes the generalization to log(n). See also math.stackexchange.com/questions/46378/… and math.stackexchange.com/questions/883348/series-for-logarithms
    – Jaume Oliver Lafont
    Jan 12 '16 at 22:20













up vote
6
down vote

favorite
6









up vote
6
down vote

favorite
6






6





Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $ln(1+x)$ and $arctan(x)$:
$$1+frac12-frac23+frac14+frac15-frac26+cdots$$



The question isn't homework or anything, just a thought tease.
I tried for a long while but couldn't find anything remotely close.
Thanks in advance for the help.










share|cite|improve this question















Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $ln(1+x)$ and $arctan(x)$:
$$1+frac12-frac23+frac14+frac15-frac26+cdots$$



The question isn't homework or anything, just a thought tease.
I tried for a long while but couldn't find anything remotely close.
Thanks in advance for the help.







calculus real-analysis sequences-and-series summation closed-form






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edited Jul 16 '15 at 8:49









user91500

1




1










asked Jan 30 '14 at 13:45









Trouble

312




312








  • 3




    The partial sum of the first $3n$ terms are $H_{3n} - H_{n}$ where $H_n$ is the $n^{th}$ harmonic number, so the answer is $log 3$.
    – achille hui
    Jan 30 '14 at 13:58










  • A comment to sequence 1,1,-2,1,1,-2,... oeis.org/A061347 includes the generalization to log(n). See also math.stackexchange.com/questions/46378/… and math.stackexchange.com/questions/883348/series-for-logarithms
    – Jaume Oliver Lafont
    Jan 12 '16 at 22:20














  • 3




    The partial sum of the first $3n$ terms are $H_{3n} - H_{n}$ where $H_n$ is the $n^{th}$ harmonic number, so the answer is $log 3$.
    – achille hui
    Jan 30 '14 at 13:58










  • A comment to sequence 1,1,-2,1,1,-2,... oeis.org/A061347 includes the generalization to log(n). See also math.stackexchange.com/questions/46378/… and math.stackexchange.com/questions/883348/series-for-logarithms
    – Jaume Oliver Lafont
    Jan 12 '16 at 22:20








3




3




The partial sum of the first $3n$ terms are $H_{3n} - H_{n}$ where $H_n$ is the $n^{th}$ harmonic number, so the answer is $log 3$.
– achille hui
Jan 30 '14 at 13:58




The partial sum of the first $3n$ terms are $H_{3n} - H_{n}$ where $H_n$ is the $n^{th}$ harmonic number, so the answer is $log 3$.
– achille hui
Jan 30 '14 at 13:58












A comment to sequence 1,1,-2,1,1,-2,... oeis.org/A061347 includes the generalization to log(n). See also math.stackexchange.com/questions/46378/… and math.stackexchange.com/questions/883348/series-for-logarithms
– Jaume Oliver Lafont
Jan 12 '16 at 22:20




A comment to sequence 1,1,-2,1,1,-2,... oeis.org/A061347 includes the generalization to log(n). See also math.stackexchange.com/questions/46378/… and math.stackexchange.com/questions/883348/series-for-logarithms
– Jaume Oliver Lafont
Jan 12 '16 at 22:20










5 Answers
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Rewrite $-frac23$ as $frac13-1$, $-frac26$ as $frac16-frac12$, and in general, $-frac{2}{3n}$ as $frac1{3n}-frac1n$. Then the sum to $3n$ terms is:



$$1 + frac12 + frac13 + frac14 + cdots + frac1{3n} - left(1 + frac12 + frac13 + frac14 + cdots + frac1nright) = sum_{r=n+1}^{3n}frac1r$$



This tends to
$$int_n^{3n}frac{dx}x = log 3$$



as $n toinfty$.






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    up vote
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    Since
    $$lim_{ntoinfty}left(1+frac1 2+frac1 3+frac1 4+frac1 5+frac1 6+...+frac1 n-log(n)right)=gamma$$
    ($gamma$ is Euler constant, for more information see here.)



    we see that
    $$gamma=lim_{ntoinfty}left(color{red}{1}+color{red}{frac1 2}+color{#0000ff}{frac1 3}+color{red}{frac1 4}+color{red}{frac1 5}+color{#0000ff}{frac1 6}+...+color{red}{frac1 {3n-2}}+color{red}{frac1 {3n-1}}+color{#0000ff}{frac1 {3n}}-log(3n)right)$$
    and
    $$gamma=lim_{ntoinfty}color{#0000ff}{3left(frac1 3+frac1 6+...+frac1 {3n}right)}-log(n)$$
    Subtracting, we obtain
    $$0=lim_{ntoinfty}left(1+frac1 2-frac2 3+frac1 4+frac1 5-frac2 6++-...+frac1{3n-2}+frac1{3n-1}-frac2 {3n}right)-log(3)$$
    which implies that
    $$1 + frac12 - frac23 + frac14 + frac15 - frac26 + ...=log(3)$$






    share|cite|improve this answer






























      up vote
      4
      down vote













      This is a direct proof using $$int_0^1 x^n dx= frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.



      $$begin{align}
      S&=sum_{k=0}^infty left(frac{1}{3k+1}+frac{1}{3k+2}-frac{2}{3k+3}right)\
      &=sum_{k=0}^infty int_0^1 left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
      &=int_0^1 sum_{k=0}^infty left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
      &=int_0^1 frac{1+x-2x^2}{1-x^3}dx\
      &=int_0^1 frac{1+2x}{1+x+x^2}dx\
      &=log(1+x+x^2)|_{0}^1\
      &=log(3)
      end{align}$$



      Thus, similarly to



      $$log(1+x)=1-frac{1}{2}x+frac{1}{3}x^2-frac{1}{4}x^3...$$



      we have



      $$log(1+x+x^2)=1+frac{1}{2}x-frac{2}{3}x^2+frac{1}{4}x^3+frac{1}{5}x^4-frac{2}{6}x^5+...$$



      Moreover, the pattern generalizes.






      share|cite|improve this answer




























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        Alternatively, note that
        $$S:=1+frac{1}{2}-frac{2}{3}+frac{1}{4}+frac{1}{5}-frac{2}{6}+ldots=sum_{k=1}^infty,frac{-omega^{k}-bar{omega}^{k}}{k},,$$
        where
        $$omega:=expleft(+frac{2pitext{i}}{3}right)=frac{-1+sqrt{3}text{i}}{2},,text{ so }bar{omega}=expleft(-frac{2pitext{i}}{3}right)=frac{-1-sqrt{3}text{i}}{2},.$$
        Since the logarithmic power series $ln(1-z)=displaystyle- sum_{k=1}^infty,frac{z^k}{k}$ converges for all $zinmathbb{C}setminus{1}$ such that $|z|leq 1$, we get
        $$S=ln(1-omega)+ln(1-bar{omega})=left(frac{1}{2},ln(3)-text{i},frac{pi}{6}right)+left(frac{1}{2},ln(3)+text{i},frac{pi}{6}right),,$$
        making $S=ln(3)$.






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          up vote
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          As every third term is "annoying", let us restore a know pattern,



          $$S:=1+frac12-frac23+frac14+frac15-frac26+cdots=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(frac33+frac36cdotsright).$$



          It is tempting to simplify and deduce



          $$S=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(1+frac12cdotsright)=color{red}{0},$$



          but this would be wrong as the sum is not absolutely summable and changing the order of the terms is not allowed.



          The right way (given by others) is by evaluating the partial sums with



          $$S_{3n}=H_{3n}-H_n$$whicht tends to$$(ln(3n)-gamma)-(ln(n)-gamma)=color{green}{ln(3)}.$$





          The reason why the simplification doesn't work is that the terms of the second series are more "sparse". In the partial sum approach, the sparsity is respected.






          share|cite|improve this answer























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            5 Answers
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            5 Answers
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            up vote
            7
            down vote













            Rewrite $-frac23$ as $frac13-1$, $-frac26$ as $frac16-frac12$, and in general, $-frac{2}{3n}$ as $frac1{3n}-frac1n$. Then the sum to $3n$ terms is:



            $$1 + frac12 + frac13 + frac14 + cdots + frac1{3n} - left(1 + frac12 + frac13 + frac14 + cdots + frac1nright) = sum_{r=n+1}^{3n}frac1r$$



            This tends to
            $$int_n^{3n}frac{dx}x = log 3$$



            as $n toinfty$.






            share|cite|improve this answer



























              up vote
              7
              down vote













              Rewrite $-frac23$ as $frac13-1$, $-frac26$ as $frac16-frac12$, and in general, $-frac{2}{3n}$ as $frac1{3n}-frac1n$. Then the sum to $3n$ terms is:



              $$1 + frac12 + frac13 + frac14 + cdots + frac1{3n} - left(1 + frac12 + frac13 + frac14 + cdots + frac1nright) = sum_{r=n+1}^{3n}frac1r$$



              This tends to
              $$int_n^{3n}frac{dx}x = log 3$$



              as $n toinfty$.






              share|cite|improve this answer

























                up vote
                7
                down vote










                up vote
                7
                down vote









                Rewrite $-frac23$ as $frac13-1$, $-frac26$ as $frac16-frac12$, and in general, $-frac{2}{3n}$ as $frac1{3n}-frac1n$. Then the sum to $3n$ terms is:



                $$1 + frac12 + frac13 + frac14 + cdots + frac1{3n} - left(1 + frac12 + frac13 + frac14 + cdots + frac1nright) = sum_{r=n+1}^{3n}frac1r$$



                This tends to
                $$int_n^{3n}frac{dx}x = log 3$$



                as $n toinfty$.






                share|cite|improve this answer














                Rewrite $-frac23$ as $frac13-1$, $-frac26$ as $frac16-frac12$, and in general, $-frac{2}{3n}$ as $frac1{3n}-frac1n$. Then the sum to $3n$ terms is:



                $$1 + frac12 + frac13 + frac14 + cdots + frac1{3n} - left(1 + frac12 + frac13 + frac14 + cdots + frac1nright) = sum_{r=n+1}^{3n}frac1r$$



                This tends to
                $$int_n^{3n}frac{dx}x = log 3$$



                as $n toinfty$.







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Jan 30 '14 at 14:17







                user93957

















                answered Jan 30 '14 at 13:59









                TonyK

                41.2k353131




                41.2k353131






















                    up vote
                    6
                    down vote













                    Since
                    $$lim_{ntoinfty}left(1+frac1 2+frac1 3+frac1 4+frac1 5+frac1 6+...+frac1 n-log(n)right)=gamma$$
                    ($gamma$ is Euler constant, for more information see here.)



                    we see that
                    $$gamma=lim_{ntoinfty}left(color{red}{1}+color{red}{frac1 2}+color{#0000ff}{frac1 3}+color{red}{frac1 4}+color{red}{frac1 5}+color{#0000ff}{frac1 6}+...+color{red}{frac1 {3n-2}}+color{red}{frac1 {3n-1}}+color{#0000ff}{frac1 {3n}}-log(3n)right)$$
                    and
                    $$gamma=lim_{ntoinfty}color{#0000ff}{3left(frac1 3+frac1 6+...+frac1 {3n}right)}-log(n)$$
                    Subtracting, we obtain
                    $$0=lim_{ntoinfty}left(1+frac1 2-frac2 3+frac1 4+frac1 5-frac2 6++-...+frac1{3n-2}+frac1{3n-1}-frac2 {3n}right)-log(3)$$
                    which implies that
                    $$1 + frac12 - frac23 + frac14 + frac15 - frac26 + ...=log(3)$$






                    share|cite|improve this answer



























                      up vote
                      6
                      down vote













                      Since
                      $$lim_{ntoinfty}left(1+frac1 2+frac1 3+frac1 4+frac1 5+frac1 6+...+frac1 n-log(n)right)=gamma$$
                      ($gamma$ is Euler constant, for more information see here.)



                      we see that
                      $$gamma=lim_{ntoinfty}left(color{red}{1}+color{red}{frac1 2}+color{#0000ff}{frac1 3}+color{red}{frac1 4}+color{red}{frac1 5}+color{#0000ff}{frac1 6}+...+color{red}{frac1 {3n-2}}+color{red}{frac1 {3n-1}}+color{#0000ff}{frac1 {3n}}-log(3n)right)$$
                      and
                      $$gamma=lim_{ntoinfty}color{#0000ff}{3left(frac1 3+frac1 6+...+frac1 {3n}right)}-log(n)$$
                      Subtracting, we obtain
                      $$0=lim_{ntoinfty}left(1+frac1 2-frac2 3+frac1 4+frac1 5-frac2 6++-...+frac1{3n-2}+frac1{3n-1}-frac2 {3n}right)-log(3)$$
                      which implies that
                      $$1 + frac12 - frac23 + frac14 + frac15 - frac26 + ...=log(3)$$






                      share|cite|improve this answer

























                        up vote
                        6
                        down vote










                        up vote
                        6
                        down vote









                        Since
                        $$lim_{ntoinfty}left(1+frac1 2+frac1 3+frac1 4+frac1 5+frac1 6+...+frac1 n-log(n)right)=gamma$$
                        ($gamma$ is Euler constant, for more information see here.)



                        we see that
                        $$gamma=lim_{ntoinfty}left(color{red}{1}+color{red}{frac1 2}+color{#0000ff}{frac1 3}+color{red}{frac1 4}+color{red}{frac1 5}+color{#0000ff}{frac1 6}+...+color{red}{frac1 {3n-2}}+color{red}{frac1 {3n-1}}+color{#0000ff}{frac1 {3n}}-log(3n)right)$$
                        and
                        $$gamma=lim_{ntoinfty}color{#0000ff}{3left(frac1 3+frac1 6+...+frac1 {3n}right)}-log(n)$$
                        Subtracting, we obtain
                        $$0=lim_{ntoinfty}left(1+frac1 2-frac2 3+frac1 4+frac1 5-frac2 6++-...+frac1{3n-2}+frac1{3n-1}-frac2 {3n}right)-log(3)$$
                        which implies that
                        $$1 + frac12 - frac23 + frac14 + frac15 - frac26 + ...=log(3)$$






                        share|cite|improve this answer














                        Since
                        $$lim_{ntoinfty}left(1+frac1 2+frac1 3+frac1 4+frac1 5+frac1 6+...+frac1 n-log(n)right)=gamma$$
                        ($gamma$ is Euler constant, for more information see here.)



                        we see that
                        $$gamma=lim_{ntoinfty}left(color{red}{1}+color{red}{frac1 2}+color{#0000ff}{frac1 3}+color{red}{frac1 4}+color{red}{frac1 5}+color{#0000ff}{frac1 6}+...+color{red}{frac1 {3n-2}}+color{red}{frac1 {3n-1}}+color{#0000ff}{frac1 {3n}}-log(3n)right)$$
                        and
                        $$gamma=lim_{ntoinfty}color{#0000ff}{3left(frac1 3+frac1 6+...+frac1 {3n}right)}-log(n)$$
                        Subtracting, we obtain
                        $$0=lim_{ntoinfty}left(1+frac1 2-frac2 3+frac1 4+frac1 5-frac2 6++-...+frac1{3n-2}+frac1{3n-1}-frac2 {3n}right)-log(3)$$
                        which implies that
                        $$1 + frac12 - frac23 + frac14 + frac15 - frac26 + ...=log(3)$$







                        share|cite|improve this answer














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                        share|cite|improve this answer








                        edited Jan 30 '14 at 17:33

























                        answered Jan 30 '14 at 14:25









                        user91500

                        1




                        1






















                            up vote
                            4
                            down vote













                            This is a direct proof using $$int_0^1 x^n dx= frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.



                            $$begin{align}
                            S&=sum_{k=0}^infty left(frac{1}{3k+1}+frac{1}{3k+2}-frac{2}{3k+3}right)\
                            &=sum_{k=0}^infty int_0^1 left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                            &=int_0^1 sum_{k=0}^infty left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                            &=int_0^1 frac{1+x-2x^2}{1-x^3}dx\
                            &=int_0^1 frac{1+2x}{1+x+x^2}dx\
                            &=log(1+x+x^2)|_{0}^1\
                            &=log(3)
                            end{align}$$



                            Thus, similarly to



                            $$log(1+x)=1-frac{1}{2}x+frac{1}{3}x^2-frac{1}{4}x^3...$$



                            we have



                            $$log(1+x+x^2)=1+frac{1}{2}x-frac{2}{3}x^2+frac{1}{4}x^3+frac{1}{5}x^4-frac{2}{6}x^5+...$$



                            Moreover, the pattern generalizes.






                            share|cite|improve this answer

























                              up vote
                              4
                              down vote













                              This is a direct proof using $$int_0^1 x^n dx= frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.



                              $$begin{align}
                              S&=sum_{k=0}^infty left(frac{1}{3k+1}+frac{1}{3k+2}-frac{2}{3k+3}right)\
                              &=sum_{k=0}^infty int_0^1 left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                              &=int_0^1 sum_{k=0}^infty left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                              &=int_0^1 frac{1+x-2x^2}{1-x^3}dx\
                              &=int_0^1 frac{1+2x}{1+x+x^2}dx\
                              &=log(1+x+x^2)|_{0}^1\
                              &=log(3)
                              end{align}$$



                              Thus, similarly to



                              $$log(1+x)=1-frac{1}{2}x+frac{1}{3}x^2-frac{1}{4}x^3...$$



                              we have



                              $$log(1+x+x^2)=1+frac{1}{2}x-frac{2}{3}x^2+frac{1}{4}x^3+frac{1}{5}x^4-frac{2}{6}x^5+...$$



                              Moreover, the pattern generalizes.






                              share|cite|improve this answer























                                up vote
                                4
                                down vote










                                up vote
                                4
                                down vote









                                This is a direct proof using $$int_0^1 x^n dx= frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.



                                $$begin{align}
                                S&=sum_{k=0}^infty left(frac{1}{3k+1}+frac{1}{3k+2}-frac{2}{3k+3}right)\
                                &=sum_{k=0}^infty int_0^1 left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                                &=int_0^1 sum_{k=0}^infty left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                                &=int_0^1 frac{1+x-2x^2}{1-x^3}dx\
                                &=int_0^1 frac{1+2x}{1+x+x^2}dx\
                                &=log(1+x+x^2)|_{0}^1\
                                &=log(3)
                                end{align}$$



                                Thus, similarly to



                                $$log(1+x)=1-frac{1}{2}x+frac{1}{3}x^2-frac{1}{4}x^3...$$



                                we have



                                $$log(1+x+x^2)=1+frac{1}{2}x-frac{2}{3}x^2+frac{1}{4}x^3+frac{1}{5}x^4-frac{2}{6}x^5+...$$



                                Moreover, the pattern generalizes.






                                share|cite|improve this answer












                                This is a direct proof using $$int_0^1 x^n dx= frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.



                                $$begin{align}
                                S&=sum_{k=0}^infty left(frac{1}{3k+1}+frac{1}{3k+2}-frac{2}{3k+3}right)\
                                &=sum_{k=0}^infty int_0^1 left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                                &=int_0^1 sum_{k=0}^infty left(x^{3k}+x^{3k+1}-2x^{3k+2}right)dx\
                                &=int_0^1 frac{1+x-2x^2}{1-x^3}dx\
                                &=int_0^1 frac{1+2x}{1+x+x^2}dx\
                                &=log(1+x+x^2)|_{0}^1\
                                &=log(3)
                                end{align}$$



                                Thus, similarly to



                                $$log(1+x)=1-frac{1}{2}x+frac{1}{3}x^2-frac{1}{4}x^3...$$



                                we have



                                $$log(1+x+x^2)=1+frac{1}{2}x-frac{2}{3}x^2+frac{1}{4}x^3+frac{1}{5}x^4-frac{2}{6}x^5+...$$



                                Moreover, the pattern generalizes.







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                                share|cite|improve this answer










                                answered May 12 '17 at 11:26









                                Jaume Oliver Lafont

                                3,04411033




                                3,04411033






















                                    up vote
                                    3
                                    down vote













                                    Alternatively, note that
                                    $$S:=1+frac{1}{2}-frac{2}{3}+frac{1}{4}+frac{1}{5}-frac{2}{6}+ldots=sum_{k=1}^infty,frac{-omega^{k}-bar{omega}^{k}}{k},,$$
                                    where
                                    $$omega:=expleft(+frac{2pitext{i}}{3}right)=frac{-1+sqrt{3}text{i}}{2},,text{ so }bar{omega}=expleft(-frac{2pitext{i}}{3}right)=frac{-1-sqrt{3}text{i}}{2},.$$
                                    Since the logarithmic power series $ln(1-z)=displaystyle- sum_{k=1}^infty,frac{z^k}{k}$ converges for all $zinmathbb{C}setminus{1}$ such that $|z|leq 1$, we get
                                    $$S=ln(1-omega)+ln(1-bar{omega})=left(frac{1}{2},ln(3)-text{i},frac{pi}{6}right)+left(frac{1}{2},ln(3)+text{i},frac{pi}{6}right),,$$
                                    making $S=ln(3)$.






                                    share|cite|improve this answer



























                                      up vote
                                      3
                                      down vote













                                      Alternatively, note that
                                      $$S:=1+frac{1}{2}-frac{2}{3}+frac{1}{4}+frac{1}{5}-frac{2}{6}+ldots=sum_{k=1}^infty,frac{-omega^{k}-bar{omega}^{k}}{k},,$$
                                      where
                                      $$omega:=expleft(+frac{2pitext{i}}{3}right)=frac{-1+sqrt{3}text{i}}{2},,text{ so }bar{omega}=expleft(-frac{2pitext{i}}{3}right)=frac{-1-sqrt{3}text{i}}{2},.$$
                                      Since the logarithmic power series $ln(1-z)=displaystyle- sum_{k=1}^infty,frac{z^k}{k}$ converges for all $zinmathbb{C}setminus{1}$ such that $|z|leq 1$, we get
                                      $$S=ln(1-omega)+ln(1-bar{omega})=left(frac{1}{2},ln(3)-text{i},frac{pi}{6}right)+left(frac{1}{2},ln(3)+text{i},frac{pi}{6}right),,$$
                                      making $S=ln(3)$.






                                      share|cite|improve this answer

























                                        up vote
                                        3
                                        down vote










                                        up vote
                                        3
                                        down vote









                                        Alternatively, note that
                                        $$S:=1+frac{1}{2}-frac{2}{3}+frac{1}{4}+frac{1}{5}-frac{2}{6}+ldots=sum_{k=1}^infty,frac{-omega^{k}-bar{omega}^{k}}{k},,$$
                                        where
                                        $$omega:=expleft(+frac{2pitext{i}}{3}right)=frac{-1+sqrt{3}text{i}}{2},,text{ so }bar{omega}=expleft(-frac{2pitext{i}}{3}right)=frac{-1-sqrt{3}text{i}}{2},.$$
                                        Since the logarithmic power series $ln(1-z)=displaystyle- sum_{k=1}^infty,frac{z^k}{k}$ converges for all $zinmathbb{C}setminus{1}$ such that $|z|leq 1$, we get
                                        $$S=ln(1-omega)+ln(1-bar{omega})=left(frac{1}{2},ln(3)-text{i},frac{pi}{6}right)+left(frac{1}{2},ln(3)+text{i},frac{pi}{6}right),,$$
                                        making $S=ln(3)$.






                                        share|cite|improve this answer














                                        Alternatively, note that
                                        $$S:=1+frac{1}{2}-frac{2}{3}+frac{1}{4}+frac{1}{5}-frac{2}{6}+ldots=sum_{k=1}^infty,frac{-omega^{k}-bar{omega}^{k}}{k},,$$
                                        where
                                        $$omega:=expleft(+frac{2pitext{i}}{3}right)=frac{-1+sqrt{3}text{i}}{2},,text{ so }bar{omega}=expleft(-frac{2pitext{i}}{3}right)=frac{-1-sqrt{3}text{i}}{2},.$$
                                        Since the logarithmic power series $ln(1-z)=displaystyle- sum_{k=1}^infty,frac{z^k}{k}$ converges for all $zinmathbb{C}setminus{1}$ such that $|z|leq 1$, we get
                                        $$S=ln(1-omega)+ln(1-bar{omega})=left(frac{1}{2},ln(3)-text{i},frac{pi}{6}right)+left(frac{1}{2},ln(3)+text{i},frac{pi}{6}right),,$$
                                        making $S=ln(3)$.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Nov 22 at 18:31

























                                        answered Sep 14 at 14:54









                                        Batominovski

                                        33.4k33292




                                        33.4k33292






















                                            up vote
                                            2
                                            down vote













                                            As every third term is "annoying", let us restore a know pattern,



                                            $$S:=1+frac12-frac23+frac14+frac15-frac26+cdots=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(frac33+frac36cdotsright).$$



                                            It is tempting to simplify and deduce



                                            $$S=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(1+frac12cdotsright)=color{red}{0},$$



                                            but this would be wrong as the sum is not absolutely summable and changing the order of the terms is not allowed.



                                            The right way (given by others) is by evaluating the partial sums with



                                            $$S_{3n}=H_{3n}-H_n$$whicht tends to$$(ln(3n)-gamma)-(ln(n)-gamma)=color{green}{ln(3)}.$$





                                            The reason why the simplification doesn't work is that the terms of the second series are more "sparse". In the partial sum approach, the sparsity is respected.






                                            share|cite|improve this answer



























                                              up vote
                                              2
                                              down vote













                                              As every third term is "annoying", let us restore a know pattern,



                                              $$S:=1+frac12-frac23+frac14+frac15-frac26+cdots=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(frac33+frac36cdotsright).$$



                                              It is tempting to simplify and deduce



                                              $$S=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(1+frac12cdotsright)=color{red}{0},$$



                                              but this would be wrong as the sum is not absolutely summable and changing the order of the terms is not allowed.



                                              The right way (given by others) is by evaluating the partial sums with



                                              $$S_{3n}=H_{3n}-H_n$$whicht tends to$$(ln(3n)-gamma)-(ln(n)-gamma)=color{green}{ln(3)}.$$





                                              The reason why the simplification doesn't work is that the terms of the second series are more "sparse". In the partial sum approach, the sparsity is respected.






                                              share|cite|improve this answer

























                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote









                                                As every third term is "annoying", let us restore a know pattern,



                                                $$S:=1+frac12-frac23+frac14+frac15-frac26+cdots=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(frac33+frac36cdotsright).$$



                                                It is tempting to simplify and deduce



                                                $$S=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(1+frac12cdotsright)=color{red}{0},$$



                                                but this would be wrong as the sum is not absolutely summable and changing the order of the terms is not allowed.



                                                The right way (given by others) is by evaluating the partial sums with



                                                $$S_{3n}=H_{3n}-H_n$$whicht tends to$$(ln(3n)-gamma)-(ln(n)-gamma)=color{green}{ln(3)}.$$





                                                The reason why the simplification doesn't work is that the terms of the second series are more "sparse". In the partial sum approach, the sparsity is respected.






                                                share|cite|improve this answer














                                                As every third term is "annoying", let us restore a know pattern,



                                                $$S:=1+frac12-frac23+frac14+frac15-frac26+cdots=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(frac33+frac36cdotsright).$$



                                                It is tempting to simplify and deduce



                                                $$S=left(1+frac12+frac13+frac14+frac15+frac16+cdotsright)-left(1+frac12cdotsright)=color{red}{0},$$



                                                but this would be wrong as the sum is not absolutely summable and changing the order of the terms is not allowed.



                                                The right way (given by others) is by evaluating the partial sums with



                                                $$S_{3n}=H_{3n}-H_n$$whicht tends to$$(ln(3n)-gamma)-(ln(n)-gamma)=color{green}{ln(3)}.$$





                                                The reason why the simplification doesn't work is that the terms of the second series are more "sparse". In the partial sum approach, the sparsity is respected.







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Sep 14 at 13:54

























                                                answered Jul 16 '15 at 9:11









                                                Yves Daoust

                                                123k668219




                                                123k668219






























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