How does this infinite series $1-frac14+frac17-frac110+cdots$ simplify to an integral $int_0^1fracdx1+x^3$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)When can a sum and integral be interchanged?Infinite Series $1+frac12-frac23+frac14+frac15-frac26+cdots$does the infinite series $sum^infty_n=1 (-1)^n frac log(n)n$ converge?Simplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Sum of an infinite series $(1 - frac 12) + (frac 12 - frac 13) + cdots$ - not geometric series?How to find the value of this integral: $int_0^nfrac 1x dx$Does this integral $int_0^infty fracdx(1+e^x)(a+x)$ have a closed form?Does this infinite series converge or diverge?Value of this convergent series: $frac13!+frac25!+frac37!+frac49!+cdots$Calculating a series for$int_0^1 fracx^p-11+x^q dx$ by interchanging sum with integral

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How does this infinite series $1-frac14+frac17-frac110+cdots$ simplify to an integral $int_0^1fracdx1+x^3$?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)When can a sum and integral be interchanged?Infinite Series $1+frac12-frac23+frac14+frac15-frac26+cdots$does the infinite series $sum^infty_n=1 (-1)^n frac log(n)n$ converge?Simplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Sum of an infinite series $(1 - frac 12) + (frac 12 - frac 13) + cdots$ - not geometric series?How to find the value of this integral: $int_0^nfrac 1x dx$Does this integral $int_0^infty fracdx(1+e^x)(a+x)$ have a closed form?Does this infinite series converge or diverge?Value of this convergent series: $frac13!+frac25!+frac37!+frac49!+cdots$Calculating a series for$int_0^1 fracx^p-11+x^q dx$ by interchanging sum with integral










2












$begingroup$


How does the infinite series below 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.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    Apr 13 at 16:05










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    Apr 13 at 16:11















2












$begingroup$


How does the infinite series below 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.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    Apr 13 at 16:05










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    Apr 13 at 16:11













2












2








2


3



$begingroup$


How does the infinite series below 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.










share|cite|improve this question











$endgroup$




How does the infinite series below 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









HAMIDINE SOUMARE

2,570417




2,570417










asked Apr 13 at 16:03









ShreeShree

134




134







  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    Apr 13 at 16:05










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    Apr 13 at 16:11












  • 1




    $begingroup$
    en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
    $endgroup$
    – lab bhattacharjee
    Apr 13 at 16:05










  • $begingroup$
    It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
    $endgroup$
    – StubbornAtom
    Apr 13 at 16:11







1




1




$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05




$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05












$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11




$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11










3 Answers
3






active

oldest

votes


















12












$begingroup$

$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    For those who aren't clear on the first equality, it's a geometric series formula.
    $endgroup$
    – Alexis Olson
    Apr 13 at 19:35










  • $begingroup$
    Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
    $endgroup$
    – StammeringMathematician
    22 hours ago






  • 1




    $begingroup$
    @StammeringMathematician Check out this question.
    $endgroup$
    – Alexis Olson
    14 hours ago










  • $begingroup$
    @AlexisOlson Thanks a lot.
    $endgroup$
    – StammeringMathematician
    12 hours ago


















4












$begingroup$

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









$endgroup$












  • $begingroup$
    Why can we interchange $sum$ and $int$
    $endgroup$
    – StammeringMathematician
    22 hours ago










  • $begingroup$
    Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
    $endgroup$
    – FDP
    21 hours ago










  • $begingroup$
    Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
    $endgroup$
    – StammeringMathematician
    21 hours ago










  • $begingroup$
    My computation shows that series converges slowly.
    $endgroup$
    – FDP
    20 hours ago


















2












$begingroup$

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






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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12












    $begingroup$

    $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
    $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      Apr 13 at 19:35










    • $begingroup$
      Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
      $endgroup$
      – StammeringMathematician
      22 hours ago






    • 1




      $begingroup$
      @StammeringMathematician Check out this question.
      $endgroup$
      – Alexis Olson
      14 hours ago










    • $begingroup$
      @AlexisOlson Thanks a lot.
      $endgroup$
      – StammeringMathematician
      12 hours ago















    12












    $begingroup$

    $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
    $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      Apr 13 at 19:35










    • $begingroup$
      Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
      $endgroup$
      – StammeringMathematician
      22 hours ago






    • 1




      $begingroup$
      @StammeringMathematician Check out this question.
      $endgroup$
      – Alexis Olson
      14 hours ago










    • $begingroup$
      @AlexisOlson Thanks a lot.
      $endgroup$
      – StammeringMathematician
      12 hours ago













    12












    12








    12





    $begingroup$

    $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
    $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$






    share|cite|improve this answer











    $endgroup$



    $$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
    $$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered Apr 13 at 16:15









    HAMIDINE SOUMAREHAMIDINE SOUMARE

    2,570417




    2,570417











    • $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      Apr 13 at 19:35










    • $begingroup$
      Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
      $endgroup$
      – StammeringMathematician
      22 hours ago






    • 1




      $begingroup$
      @StammeringMathematician Check out this question.
      $endgroup$
      – Alexis Olson
      14 hours ago










    • $begingroup$
      @AlexisOlson Thanks a lot.
      $endgroup$
      – StammeringMathematician
      12 hours ago
















    • $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      Apr 13 at 19:35










    • $begingroup$
      Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
      $endgroup$
      – StammeringMathematician
      22 hours ago






    • 1




      $begingroup$
      @StammeringMathematician Check out this question.
      $endgroup$
      – Alexis Olson
      14 hours ago










    • $begingroup$
      @AlexisOlson Thanks a lot.
      $endgroup$
      – StammeringMathematician
      12 hours ago















    $begingroup$
    For those who aren't clear on the first equality, it's a geometric series formula.
    $endgroup$
    – Alexis Olson
    Apr 13 at 19:35




    $begingroup$
    For those who aren't clear on the first equality, it's a geometric series formula.
    $endgroup$
    – Alexis Olson
    Apr 13 at 19:35












    $begingroup$
    Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
    $endgroup$
    – StammeringMathematician
    22 hours ago




    $begingroup$
    Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
    $endgroup$
    – StammeringMathematician
    22 hours ago




    1




    1




    $begingroup$
    @StammeringMathematician Check out this question.
    $endgroup$
    – Alexis Olson
    14 hours ago




    $begingroup$
    @StammeringMathematician Check out this question.
    $endgroup$
    – Alexis Olson
    14 hours ago












    $begingroup$
    @AlexisOlson Thanks a lot.
    $endgroup$
    – StammeringMathematician
    12 hours ago




    $begingroup$
    @AlexisOlson Thanks a lot.
    $endgroup$
    – StammeringMathematician
    12 hours ago











    4












    $begingroup$

    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









    $endgroup$












    • $begingroup$
      Why can we interchange $sum$ and $int$
      $endgroup$
      – StammeringMathematician
      22 hours ago










    • $begingroup$
      Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
      $endgroup$
      – FDP
      21 hours ago










    • $begingroup$
      Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
      $endgroup$
      – StammeringMathematician
      21 hours ago










    • $begingroup$
      My computation shows that series converges slowly.
      $endgroup$
      – FDP
      20 hours ago















    4












    $begingroup$

    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









    $endgroup$












    • $begingroup$
      Why can we interchange $sum$ and $int$
      $endgroup$
      – StammeringMathematician
      22 hours ago










    • $begingroup$
      Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
      $endgroup$
      – FDP
      21 hours ago










    • $begingroup$
      Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
      $endgroup$
      – StammeringMathematician
      21 hours ago










    • $begingroup$
      My computation shows that series converges slowly.
      $endgroup$
      – FDP
      20 hours ago













    4












    4








    4





    $begingroup$

    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 Apr 13 at 16:47









    FDPFDP

    6,16211929




    6,16211929











    • $begingroup$
      Why can we interchange $sum$ and $int$
      $endgroup$
      – StammeringMathematician
      22 hours ago










    • $begingroup$
      Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
      $endgroup$
      – FDP
      21 hours ago










    • $begingroup$
      Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
      $endgroup$
      – StammeringMathematician
      21 hours ago










    • $begingroup$
      My computation shows that series converges slowly.
      $endgroup$
      – FDP
      20 hours ago
















    • $begingroup$
      Why can we interchange $sum$ and $int$
      $endgroup$
      – StammeringMathematician
      22 hours ago










    • $begingroup$
      Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
      $endgroup$
      – FDP
      21 hours ago










    • $begingroup$
      Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
      $endgroup$
      – StammeringMathematician
      21 hours ago










    • $begingroup$
      My computation shows that series converges slowly.
      $endgroup$
      – FDP
      20 hours ago















    $begingroup$
    Why can we interchange $sum$ and $int$
    $endgroup$
    – StammeringMathematician
    22 hours ago




    $begingroup$
    Why can we interchange $sum$ and $int$
    $endgroup$
    – StammeringMathematician
    22 hours ago












    $begingroup$
    Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
    $endgroup$
    – FDP
    21 hours ago




    $begingroup$
    Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
    $endgroup$
    – FDP
    21 hours ago












    $begingroup$
    Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
    $endgroup$
    – StammeringMathematician
    21 hours ago




    $begingroup$
    Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
    $endgroup$
    – StammeringMathematician
    21 hours ago












    $begingroup$
    My computation shows that series converges slowly.
    $endgroup$
    – FDP
    20 hours ago




    $begingroup$
    My computation shows that series converges slowly.
    $endgroup$
    – FDP
    20 hours ago











    2












    $begingroup$

    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









    $endgroup$

















      2












      $begingroup$

      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









      $endgroup$















        2












        2








        2





        $begingroup$

        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 Apr 13 at 16:15









        José Carlos SantosJosé Carlos Santos

        175k24134243




        175k24134243



























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