Prove that $limlimits_xto 1^-f(x)=-1+ln 4$ where $f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)$

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Prove that $limlimits_xto 1^-f(x)=-1+ln 4$ where $$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1).$$



My trial:



Since the series converges absolutely on $[-1,1]$, then it is uniformly convergent on $[-1,1].$ So,



$$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)$$
$$=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]$$
$$=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=1^infty(-1)^nfracx^nn-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=0^infty(-1)^n+1fracx^n+1n+1-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=0^infty(-1)^n+1fracx^n+1n+1-x+xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=1-Big(frac1x+1Big)ln(1+x)$$
So, $$limlimits_xto 1^-f(x)=1-ln 4.$$
Please, where am I missing it? Better solutions will be accepted.




real-analysis sequences-and-series analysis limits continuity




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    "Better solutions will be accepted", hurray ;)
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Prove that $limlimits_xto 1^-f(x)=-1+ln 4$ where $$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1).$$



My trial:



Since the series converges absolutely on $[-1,1]$, then it is uniformly convergent on $[-1,1].$ So,



$$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)$$
$$=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]$$
$$=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=1^infty(-1)^nfracx^nn-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=0^infty(-1)^n+1fracx^n+1n+1-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=0^infty(-1)^n+1fracx^n+1n+1-x+xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=1-Big(frac1x+1Big)ln(1+x)$$
So, $$limlimits_xto 1^-f(x)=1-ln 4.$$
Please, where am I missing it? Better solutions will be accepted.




real-analysis sequences-and-series analysis limits continuity




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  • 2




    "Better solutions will be accepted", hurray ;)
    – Nosrati
    Jul 18 at 8:01












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Prove that $limlimits_xto 1^-f(x)=-1+ln 4$ where $$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1).$$



My trial:



Since the series converges absolutely on $[-1,1]$, then it is uniformly convergent on $[-1,1].$ So,



$$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)$$
$$=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]$$
$$=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=1^infty(-1)^nfracx^nn-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=0^infty(-1)^n+1fracx^n+1n+1-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=0^infty(-1)^n+1fracx^n+1n+1-x+xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=1-Big(frac1x+1Big)ln(1+x)$$
So, $$limlimits_xto 1^-f(x)=1-ln 4.$$
Please, where am I missing it? Better solutions will be accepted.




real-analysis sequences-and-series analysis limits continuity




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Prove that $limlimits_xto 1^-f(x)=-1+ln 4$ where $$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1).$$



My trial:



Since the series converges absolutely on $[-1,1]$, then it is uniformly convergent on $[-1,1].$ So,



$$f(x)=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)$$
$$=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]$$
$$=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=1^infty(-1)^nfracx^nn-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xsum_n=0^infty(-1)^n+1fracx^n+1n+1-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=0^infty(-1)^n+1fracx^n+1n+1-x+xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=-frac1xBig[sum_n=1^infty(-1)^n+1fracx^n+1n+1-xBig]-sum_n=1^infty(-1)^n-1fracx^n+1n+1$$
$$=1-Big(frac1x+1Big)ln(1+x)$$
So, $$limlimits_xto 1^-f(x)=1-ln 4.$$
Please, where am I missing it? Better solutions will be accepted.





real-analysis sequences-and-series analysis limits continuity





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asked Jul 18 at 7:31









Mike

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  • 2




    "Better solutions will be accepted", hurray ;)
    – Nosrati
    Jul 18 at 8:01












  • 2




    "Better solutions will be accepted", hurray ;)
    – Nosrati
    Jul 18 at 8:01







2




2




"Better solutions will be accepted", hurray ;)
– Nosrati
Jul 18 at 8:01




"Better solutions will be accepted", hurray ;)
– Nosrati
Jul 18 at 8:01










4 Answers
4






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up vote
2
down vote



accepted










I rewrite part of yours:
beginalign
f(x)
&=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)\
&=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]\
&=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1\
&=xsum_n=1^infty(-1)^n-1fracx^nn+sum_n=1^infty(-1)^nfracx^n+1n+1+x-x\
&=xleft(sum_n=1^infty(-1)^n-1fracx^nnright)+left(sum_n=1^infty(-1)^n-1fracx^nnright)-x\
&=(1+x)ln(1+x)-x
endalign






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  • Oh, my bad! So, I should have done this instead! Thanks always!
    – Mike
    Jul 18 at 9:44

















up vote
1
down vote













$ln (1+x)=x-x^2/2+... $ but you are writing $ln (1+x)=x^2/2-x^3/3... $. All your calculations are fine except for this mistake at the end. BTW you can take the limit inside the sum right in the beginning (and get rid of $x$) because of uniform convergence).






share|cite|improve this answer






























    up vote
    1
    down vote













    It's not hard to see that $f$ is infinitely many times differentiable in $(-1,1)$ (use ratio test) therefore $$f''(x)=sum_n=1^infty(-x)^n-1=dfrac11+x$$therefore $$f'(x)=ln(1+x)+C_1$$and $$f(x)=(1+x)ln(1+x)-1-x+C_1x+C_2$$and we know that $$C_1=C_2=0$$therefore $$lim_xto1^-f(x)=2ln2-1$$






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      A Different Approach
      $$
      beginalign
      lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
      &=lim_xto1^-sum_n=1^inftyfracx^2n(2n-1)2n
      -lim_xto1^-sum_n=1^inftyfracx^2n+12n(2n+1)tag1\
      &=sum_n=1^inftyfrac1(2n-1)2n
      -sum_n=1^inftyfrac12n(2n+1)tag2\
      &=sum_n=1^inftyleft(frac12n-1-frac22n+frac12n+1right)tag3\
      &=sum_n=1^inftyleft(frac22n-1-frac22nright)-sum_n=1^inftyleft(frac12n-1-frac12n+1right)tag4\[9pt]
      &=2log(2)-1tag5
      endalign
      $$
      Explanation:

      $(1)$: separate the series into the difference of the even and the odd terms

      $(2)$: apply monotone or dominated convergence to both sums (either works)

      $(3)$: partial fractions

      $(4)$: separate the series into the difference of two convergent series

      $(5)$: the series on the left is twice the alternating harmonic series

      $phantom(5)text:$ the series on the right telescopes



      A Modification of the Approach in the Question
      $$
      beginalign
      lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
      &=lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n-lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n+1tag6\
      &=lim_xto1^-xlog(1+x)-lim_xto1^-(1-log(1+x))tag7\[6pt]
      &=2log(2)-1tag8
      endalign
      $$
      Explanation:

      $(6)$: partial fractions, then separate into two convergent series

      $(7)$: the series converge absolutely to $log(1+x)$ for $xlt1$

      $(8)$: evaluate at $x=1$






      share|cite|improve this answer























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        4 Answers
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        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        I rewrite part of yours:
        beginalign
        f(x)
        &=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)\
        &=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]\
        &=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1\
        &=xsum_n=1^infty(-1)^n-1fracx^nn+sum_n=1^infty(-1)^nfracx^n+1n+1+x-x\
        &=xleft(sum_n=1^infty(-1)^n-1fracx^nnright)+left(sum_n=1^infty(-1)^n-1fracx^nnright)-x\
        &=(1+x)ln(1+x)-x
        endalign






        share|cite|improve this answer





















        • Oh, my bad! So, I should have done this instead! Thanks always!
          – Mike
          Jul 18 at 9:44














        up vote
        2
        down vote



        accepted










        I rewrite part of yours:
        beginalign
        f(x)
        &=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)\
        &=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]\
        &=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1\
        &=xsum_n=1^infty(-1)^n-1fracx^nn+sum_n=1^infty(-1)^nfracx^n+1n+1+x-x\
        &=xleft(sum_n=1^infty(-1)^n-1fracx^nnright)+left(sum_n=1^infty(-1)^n-1fracx^nnright)-x\
        &=(1+x)ln(1+x)-x
        endalign






        share|cite|improve this answer





















        • Oh, my bad! So, I should have done this instead! Thanks always!
          – Mike
          Jul 18 at 9:44












        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        I rewrite part of yours:
        beginalign
        f(x)
        &=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)\
        &=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]\
        &=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1\
        &=xsum_n=1^infty(-1)^n-1fracx^nn+sum_n=1^infty(-1)^nfracx^n+1n+1+x-x\
        &=xleft(sum_n=1^infty(-1)^n-1fracx^nnright)+left(sum_n=1^infty(-1)^n-1fracx^nnright)-x\
        &=(1+x)ln(1+x)-x
        endalign






        share|cite|improve this answer













        I rewrite part of yours:
        beginalign
        f(x)
        &=sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)\
        &=sum_n=1^infty(-1)^n-1x^n+1Big[frac1n-frac1n+1Big]\
        &=sum_n=1^infty(-1)^n-1fracx^n+1n-sum_n=1^infty(-1)^n-1fracx^n+1n+1\
        &=xsum_n=1^infty(-1)^n-1fracx^nn+sum_n=1^infty(-1)^nfracx^n+1n+1+x-x\
        &=xleft(sum_n=1^infty(-1)^n-1fracx^nnright)+left(sum_n=1^infty(-1)^n-1fracx^nnright)-x\
        &=(1+x)ln(1+x)-x
        endalign







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        answered Jul 18 at 8:14









        Nosrati

        19.7k41644




        19.7k41644











        • Oh, my bad! So, I should have done this instead! Thanks always!
          – Mike
          Jul 18 at 9:44
















        • Oh, my bad! So, I should have done this instead! Thanks always!
          – Mike
          Jul 18 at 9:44















        Oh, my bad! So, I should have done this instead! Thanks always!
        – Mike
        Jul 18 at 9:44




        Oh, my bad! So, I should have done this instead! Thanks always!
        – Mike
        Jul 18 at 9:44










        up vote
        1
        down vote













        $ln (1+x)=x-x^2/2+... $ but you are writing $ln (1+x)=x^2/2-x^3/3... $. All your calculations are fine except for this mistake at the end. BTW you can take the limit inside the sum right in the beginning (and get rid of $x$) because of uniform convergence).






        share|cite|improve this answer



























          up vote
          1
          down vote













          $ln (1+x)=x-x^2/2+... $ but you are writing $ln (1+x)=x^2/2-x^3/3... $. All your calculations are fine except for this mistake at the end. BTW you can take the limit inside the sum right in the beginning (and get rid of $x$) because of uniform convergence).






          share|cite|improve this answer

























            up vote
            1
            down vote










            up vote
            1
            down vote









            $ln (1+x)=x-x^2/2+... $ but you are writing $ln (1+x)=x^2/2-x^3/3... $. All your calculations are fine except for this mistake at the end. BTW you can take the limit inside the sum right in the beginning (and get rid of $x$) because of uniform convergence).






            share|cite|improve this answer















            $ln (1+x)=x-x^2/2+... $ but you are writing $ln (1+x)=x^2/2-x^3/3... $. All your calculations are fine except for this mistake at the end. BTW you can take the limit inside the sum right in the beginning (and get rid of $x$) because of uniform convergence).







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 18 at 8:03


























            answered Jul 18 at 7:42









            Kavi Rama Murthy

            20.8k2829




            20.8k2829




















                up vote
                1
                down vote













                It's not hard to see that $f$ is infinitely many times differentiable in $(-1,1)$ (use ratio test) therefore $$f''(x)=sum_n=1^infty(-x)^n-1=dfrac11+x$$therefore $$f'(x)=ln(1+x)+C_1$$and $$f(x)=(1+x)ln(1+x)-1-x+C_1x+C_2$$and we know that $$C_1=C_2=0$$therefore $$lim_xto1^-f(x)=2ln2-1$$






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  It's not hard to see that $f$ is infinitely many times differentiable in $(-1,1)$ (use ratio test) therefore $$f''(x)=sum_n=1^infty(-x)^n-1=dfrac11+x$$therefore $$f'(x)=ln(1+x)+C_1$$and $$f(x)=(1+x)ln(1+x)-1-x+C_1x+C_2$$and we know that $$C_1=C_2=0$$therefore $$lim_xto1^-f(x)=2ln2-1$$






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    It's not hard to see that $f$ is infinitely many times differentiable in $(-1,1)$ (use ratio test) therefore $$f''(x)=sum_n=1^infty(-x)^n-1=dfrac11+x$$therefore $$f'(x)=ln(1+x)+C_1$$and $$f(x)=(1+x)ln(1+x)-1-x+C_1x+C_2$$and we know that $$C_1=C_2=0$$therefore $$lim_xto1^-f(x)=2ln2-1$$






                    share|cite|improve this answer













                    It's not hard to see that $f$ is infinitely many times differentiable in $(-1,1)$ (use ratio test) therefore $$f''(x)=sum_n=1^infty(-x)^n-1=dfrac11+x$$therefore $$f'(x)=ln(1+x)+C_1$$and $$f(x)=(1+x)ln(1+x)-1-x+C_1x+C_2$$and we know that $$C_1=C_2=0$$therefore $$lim_xto1^-f(x)=2ln2-1$$







                    share|cite|improve this answer













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                    answered Jul 18 at 8:35









                    Mostafa Ayaz

                    8,6023630




                    8,6023630




















                        up vote
                        1
                        down vote













                        A Different Approach
                        $$
                        beginalign
                        lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                        &=lim_xto1^-sum_n=1^inftyfracx^2n(2n-1)2n
                        -lim_xto1^-sum_n=1^inftyfracx^2n+12n(2n+1)tag1\
                        &=sum_n=1^inftyfrac1(2n-1)2n
                        -sum_n=1^inftyfrac12n(2n+1)tag2\
                        &=sum_n=1^inftyleft(frac12n-1-frac22n+frac12n+1right)tag3\
                        &=sum_n=1^inftyleft(frac22n-1-frac22nright)-sum_n=1^inftyleft(frac12n-1-frac12n+1right)tag4\[9pt]
                        &=2log(2)-1tag5
                        endalign
                        $$
                        Explanation:

                        $(1)$: separate the series into the difference of the even and the odd terms

                        $(2)$: apply monotone or dominated convergence to both sums (either works)

                        $(3)$: partial fractions

                        $(4)$: separate the series into the difference of two convergent series

                        $(5)$: the series on the left is twice the alternating harmonic series

                        $phantom(5)text:$ the series on the right telescopes



                        A Modification of the Approach in the Question
                        $$
                        beginalign
                        lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                        &=lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n-lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n+1tag6\
                        &=lim_xto1^-xlog(1+x)-lim_xto1^-(1-log(1+x))tag7\[6pt]
                        &=2log(2)-1tag8
                        endalign
                        $$
                        Explanation:

                        $(6)$: partial fractions, then separate into two convergent series

                        $(7)$: the series converge absolutely to $log(1+x)$ for $xlt1$

                        $(8)$: evaluate at $x=1$






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          A Different Approach
                          $$
                          beginalign
                          lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                          &=lim_xto1^-sum_n=1^inftyfracx^2n(2n-1)2n
                          -lim_xto1^-sum_n=1^inftyfracx^2n+12n(2n+1)tag1\
                          &=sum_n=1^inftyfrac1(2n-1)2n
                          -sum_n=1^inftyfrac12n(2n+1)tag2\
                          &=sum_n=1^inftyleft(frac12n-1-frac22n+frac12n+1right)tag3\
                          &=sum_n=1^inftyleft(frac22n-1-frac22nright)-sum_n=1^inftyleft(frac12n-1-frac12n+1right)tag4\[9pt]
                          &=2log(2)-1tag5
                          endalign
                          $$
                          Explanation:

                          $(1)$: separate the series into the difference of the even and the odd terms

                          $(2)$: apply monotone or dominated convergence to both sums (either works)

                          $(3)$: partial fractions

                          $(4)$: separate the series into the difference of two convergent series

                          $(5)$: the series on the left is twice the alternating harmonic series

                          $phantom(5)text:$ the series on the right telescopes



                          A Modification of the Approach in the Question
                          $$
                          beginalign
                          lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                          &=lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n-lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n+1tag6\
                          &=lim_xto1^-xlog(1+x)-lim_xto1^-(1-log(1+x))tag7\[6pt]
                          &=2log(2)-1tag8
                          endalign
                          $$
                          Explanation:

                          $(6)$: partial fractions, then separate into two convergent series

                          $(7)$: the series converge absolutely to $log(1+x)$ for $xlt1$

                          $(8)$: evaluate at $x=1$






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                            A Different Approach
                            $$
                            beginalign
                            lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                            &=lim_xto1^-sum_n=1^inftyfracx^2n(2n-1)2n
                            -lim_xto1^-sum_n=1^inftyfracx^2n+12n(2n+1)tag1\
                            &=sum_n=1^inftyfrac1(2n-1)2n
                            -sum_n=1^inftyfrac12n(2n+1)tag2\
                            &=sum_n=1^inftyleft(frac12n-1-frac22n+frac12n+1right)tag3\
                            &=sum_n=1^inftyleft(frac22n-1-frac22nright)-sum_n=1^inftyleft(frac12n-1-frac12n+1right)tag4\[9pt]
                            &=2log(2)-1tag5
                            endalign
                            $$
                            Explanation:

                            $(1)$: separate the series into the difference of the even and the odd terms

                            $(2)$: apply monotone or dominated convergence to both sums (either works)

                            $(3)$: partial fractions

                            $(4)$: separate the series into the difference of two convergent series

                            $(5)$: the series on the left is twice the alternating harmonic series

                            $phantom(5)text:$ the series on the right telescopes



                            A Modification of the Approach in the Question
                            $$
                            beginalign
                            lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                            &=lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n-lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n+1tag6\
                            &=lim_xto1^-xlog(1+x)-lim_xto1^-(1-log(1+x))tag7\[6pt]
                            &=2log(2)-1tag8
                            endalign
                            $$
                            Explanation:

                            $(6)$: partial fractions, then separate into two convergent series

                            $(7)$: the series converge absolutely to $log(1+x)$ for $xlt1$

                            $(8)$: evaluate at $x=1$






                            share|cite|improve this answer















                            A Different Approach
                            $$
                            beginalign
                            lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                            &=lim_xto1^-sum_n=1^inftyfracx^2n(2n-1)2n
                            -lim_xto1^-sum_n=1^inftyfracx^2n+12n(2n+1)tag1\
                            &=sum_n=1^inftyfrac1(2n-1)2n
                            -sum_n=1^inftyfrac12n(2n+1)tag2\
                            &=sum_n=1^inftyleft(frac12n-1-frac22n+frac12n+1right)tag3\
                            &=sum_n=1^inftyleft(frac22n-1-frac22nright)-sum_n=1^inftyleft(frac12n-1-frac12n+1right)tag4\[9pt]
                            &=2log(2)-1tag5
                            endalign
                            $$
                            Explanation:

                            $(1)$: separate the series into the difference of the even and the odd terms

                            $(2)$: apply monotone or dominated convergence to both sums (either works)

                            $(3)$: partial fractions

                            $(4)$: separate the series into the difference of two convergent series

                            $(5)$: the series on the left is twice the alternating harmonic series

                            $phantom(5)text:$ the series on the right telescopes



                            A Modification of the Approach in the Question
                            $$
                            beginalign
                            lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n(n+1)
                            &=lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n-lim_xto1^-sum_n=1^infty(-1)^n-1fracx^n+1n+1tag6\
                            &=lim_xto1^-xlog(1+x)-lim_xto1^-(1-log(1+x))tag7\[6pt]
                            &=2log(2)-1tag8
                            endalign
                            $$
                            Explanation:

                            $(6)$: partial fractions, then separate into two convergent series

                            $(7)$: the series converge absolutely to $log(1+x)$ for $xlt1$

                            $(8)$: evaluate at $x=1$







                            share|cite|improve this answer















                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jul 18 at 14:58


























                            answered Jul 18 at 8:25









                            robjohn♦

                            258k26297612




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