How does this infinite series simplify to an integral? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Infinite Series ManipulationsHow do I solve this integral? (looking for hints)Finding Sum for Infinite SeriesSimplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Does this infinite series converge or diverge?Sum of this infinite series…Why does Improper Integral Diverge?Infinite series containing positive and negative termsInfinite series sum which have infinite terms

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How does this infinite series simplify to an integral?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Infinite Series ManipulationsHow do I solve this integral? (looking for hints)Finding Sum for Infinite SeriesSimplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Does this infinite series converge or diverge?Sum of this infinite series…Why does Improper Integral Diverge?Infinite series containing positive and negative termsInfinite series sum which have infinite terms










2












$begingroup$


How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    3 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    3 hours ago















2












$begingroup$


How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    3 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    3 hours ago













2












2








2


0



$begingroup$


How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals







share|cite|improve this question











$endgroup$




How does the infinite series in the link above simplify to that integral?



$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$



I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.






integration sequences-and-series power-series improper-integrals




integration sequences-and-series power-series improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Zacky

7,87511062




7,87511062










asked 3 hours ago









ShreeShree

114




114







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    3 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    3 hours ago












  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    3 hours ago










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    3 hours ago







1




1




$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
3 hours ago




$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
3 hours ago












$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
3 hours ago




$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
3 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

$$int_0^1frac11-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

      for $x$ real, $ngeq 0$ integer
      beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
      &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
      endalign



      For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



      Therefore,
      beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
      &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
      &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
      endalign



      For $xin[0;1],ngeq 0$, integer,
      beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
      and,
      beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
      Therefore,
      beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
      beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
      Therefore,
      beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        3












        $begingroup$

        $$int_0^1frac11-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
        $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






        share|cite|improve this answer









        $endgroup$

















          3












          $begingroup$

          $$int_0^1frac11-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
          $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






          share|cite|improve this answer









          $endgroup$















            3












            3








            3





            $begingroup$

            $$int_0^1frac11-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
            $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






            share|cite|improve this answer









            $endgroup$



            $$int_0^1frac11-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
            $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            HAMIDINE SOUMAREHAMIDINE SOUMARE

            2,156212




            2,156212





















                1












                $begingroup$

                If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign






                    share|cite|improve this answer









                    $endgroup$



                    If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    José Carlos SantosJosé Carlos Santos

                    174k23133243




                    174k23133243





















                        1












                        $begingroup$

                        for $x$ real, $ngeq 0$ integer
                        beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
                        &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
                        endalign



                        For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



                        Therefore,
                        beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
                        &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
                        &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
                        endalign



                        For $xin[0;1],ngeq 0$, integer,
                        beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
                        and,
                        beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
                        Therefore,
                        beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
                        beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
                        Therefore,
                        beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          for $x$ real, $ngeq 0$ integer
                          beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
                          &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
                          endalign



                          For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



                          Therefore,
                          beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
                          &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
                          &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
                          endalign



                          For $xin[0;1],ngeq 0$, integer,
                          beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
                          and,
                          beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
                          Therefore,
                          beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
                          beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
                          Therefore,
                          beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            for $x$ real, $ngeq 0$ integer
                            beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
                            &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
                            endalign



                            For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



                            Therefore,
                            beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
                            &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
                            &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
                            endalign



                            For $xin[0;1],ngeq 0$, integer,
                            beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
                            and,
                            beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
                            Therefore,
                            beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
                            beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
                            Therefore,
                            beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign






                            share|cite|improve this answer









                            $endgroup$



                            for $x$ real, $ngeq 0$ integer
                            beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
                            &=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
                            endalign



                            For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign



                            Therefore,
                            beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
                            &=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
                            &=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
                            endalign



                            For $xin[0;1],ngeq 0$, integer,
                            beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
                            and,
                            beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
                            Therefore,
                            beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
                            beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
                            Therefore,
                            beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            FDPFDP

                            6,11211929




                            6,11211929



























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