How does this infinite series simplify to an integral?












2












$begingroup$


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



$$1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots=int_0^1frac{dx}{1+x^3}$$



I thought of simplifying the series to the sum to infinity of $frac{1}{6n-5} - frac{1}{6n-2}$, but this did not help.






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







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  • 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?



$$1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots=int_0^1frac{dx}{1+x^3}$$



I thought of simplifying the series to the sum to infinity of $frac{1}{6n-5} - frac{1}{6n-2}$, 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?



$$1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots=int_0^1frac{dx}{1+x^3}$$



I thought of simplifying the series to the sum to infinity of $frac{1}{6n-5} - frac{1}{6n-2}$, 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?



$$1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots=int_0^1frac{dx}{1+x^3}$$



I thought of simplifying the series to the sum to infinity of $frac{1}{6n-5} - frac{1}{6n-2}$, 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













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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






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oldest

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5












$begingroup$

$$int_{0}^{1}{frac{1}{1-(-x)^3}=int_{0}^{1}{sum_{n=0}^{infty}{(-x)}^{3n}}}dx=sum_{n=0}^{infty}{(-1)^{3n}int_{0}^{1}{x^{3n}}dx}$$
$$=sum_{n=0}^{infty}{frac{(-1)^{3n}}{3n+1}}= 1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots $$






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
    18 mins ago



















2












$begingroup$

for $x$ real, $ngeq 0$ integer
begin{align}frac{1}{1+x^3}&=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1-(-x^3)}\
&=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1+x^3}\
end{align}



For $xneq 1$, $ngeq 0$ integer, begin{align}sum_{k=0}^n x^k=frac{1-x^{n+1}}{1-x}end{align}



Therefore,
begin{align}int_0^1 frac{1}{1+x^3},dx&=int_0^1 left(sum_{k=0}^n (-x^3)^kright),dx+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
&=sum_{k=0}^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
&=sum_{k=0}^n frac{(-1)^k}{3k+1}+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
end{align}



For $xin[0;1],ngeq 0$, integer,
begin{align}frac{x^{3(n+1)}}{1+x^3}leq x^{3(n+1)}end{align}
and,
begin{align}int_0^1 x^{3(n+1)},dx=frac{1}{3n+4}end{align}
Therefore,
begin{align}left|int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dxright|leq frac{1}{3n+4}end{align}
begin{align}left|int_0^1 frac{1}{1+x^3},dx-sum_{k=0}^n frac{(-1)^k}{3k+1}right|leq frac{1}{3n+4}end{align}
Therefore,
begin{align}boxed{int_0^1 frac{1}{1+x^3},dx=sum_{k=0}^infty frac{(-1)^k}{3k+1}}end{align}






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$endgroup$





















    1












    $begingroup$

    If $lvert xrvert<1$, let$$f(x)=sum_{n=0}^inftyfrac{x^{3n+1}}{3n+1}.$$Then $$f'(x)=sum_{n=0}^infty x^{3n}=frac1{1-x^3}.$$Thereforebegin{align}1-frac14+frac17-frac1{10}+cdots&=lim_{xto1}f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac1{1-x^3},mathrm dx.end{align}






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

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      active

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      active

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      5












      $begingroup$

      $$int_{0}^{1}{frac{1}{1-(-x)^3}=int_{0}^{1}{sum_{n=0}^{infty}{(-x)}^{3n}}}dx=sum_{n=0}^{infty}{(-1)^{3n}int_{0}^{1}{x^{3n}}dx}$$
      $$=sum_{n=0}^{infty}{frac{(-1)^{3n}}{3n+1}}= 1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots $$






      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
        18 mins ago
















      5












      $begingroup$

      $$int_{0}^{1}{frac{1}{1-(-x)^3}=int_{0}^{1}{sum_{n=0}^{infty}{(-x)}^{3n}}}dx=sum_{n=0}^{infty}{(-1)^{3n}int_{0}^{1}{x^{3n}}dx}$$
      $$=sum_{n=0}^{infty}{frac{(-1)^{3n}}{3n+1}}= 1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots $$






      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
        18 mins ago














      5












      5








      5





      $begingroup$

      $$int_{0}^{1}{frac{1}{1-(-x)^3}=int_{0}^{1}{sum_{n=0}^{infty}{(-x)}^{3n}}}dx=sum_{n=0}^{infty}{(-1)^{3n}int_{0}^{1}{x^{3n}}dx}$$
      $$=sum_{n=0}^{infty}{frac{(-1)^{3n}}{3n+1}}= 1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots $$






      share|cite|improve this answer









      $endgroup$



      $$int_{0}^{1}{frac{1}{1-(-x)^3}=int_{0}^{1}{sum_{n=0}^{infty}{(-x)}^{3n}}}dx=sum_{n=0}^{infty}{(-1)^{3n}int_{0}^{1}{x^{3n}}dx}$$
      $$=sum_{n=0}^{infty}{frac{(-1)^{3n}}{3n+1}}= 1-frac{1}{4}+frac{1}{7}-frac{1}{10}+cdots $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 3 hours ago









      HAMIDINE SOUMAREHAMIDINE SOUMARE

      2,184212




      2,184212












      • $begingroup$
        For those who aren't clear on the first equality, it's a geometric series formula.
        $endgroup$
        – Alexis Olson
        18 mins ago


















      • $begingroup$
        For those who aren't clear on the first equality, it's a geometric series formula.
        $endgroup$
        – Alexis Olson
        18 mins ago
















      $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      18 mins ago




      $begingroup$
      For those who aren't clear on the first equality, it's a geometric series formula.
      $endgroup$
      – Alexis Olson
      18 mins ago











      2












      $begingroup$

      for $x$ real, $ngeq 0$ integer
      begin{align}frac{1}{1+x^3}&=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1-(-x^3)}\
      &=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1+x^3}\
      end{align}



      For $xneq 1$, $ngeq 0$ integer, begin{align}sum_{k=0}^n x^k=frac{1-x^{n+1}}{1-x}end{align}



      Therefore,
      begin{align}int_0^1 frac{1}{1+x^3},dx&=int_0^1 left(sum_{k=0}^n (-x^3)^kright),dx+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
      &=sum_{k=0}^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
      &=sum_{k=0}^n frac{(-1)^k}{3k+1}+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
      end{align}



      For $xin[0;1],ngeq 0$, integer,
      begin{align}frac{x^{3(n+1)}}{1+x^3}leq x^{3(n+1)}end{align}
      and,
      begin{align}int_0^1 x^{3(n+1)},dx=frac{1}{3n+4}end{align}
      Therefore,
      begin{align}left|int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dxright|leq frac{1}{3n+4}end{align}
      begin{align}left|int_0^1 frac{1}{1+x^3},dx-sum_{k=0}^n frac{(-1)^k}{3k+1}right|leq frac{1}{3n+4}end{align}
      Therefore,
      begin{align}boxed{int_0^1 frac{1}{1+x^3},dx=sum_{k=0}^infty frac{(-1)^k}{3k+1}}end{align}






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        for $x$ real, $ngeq 0$ integer
        begin{align}frac{1}{1+x^3}&=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1-(-x^3)}\
        &=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1+x^3}\
        end{align}



        For $xneq 1$, $ngeq 0$ integer, begin{align}sum_{k=0}^n x^k=frac{1-x^{n+1}}{1-x}end{align}



        Therefore,
        begin{align}int_0^1 frac{1}{1+x^3},dx&=int_0^1 left(sum_{k=0}^n (-x^3)^kright),dx+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
        &=sum_{k=0}^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
        &=sum_{k=0}^n frac{(-1)^k}{3k+1}+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
        end{align}



        For $xin[0;1],ngeq 0$, integer,
        begin{align}frac{x^{3(n+1)}}{1+x^3}leq x^{3(n+1)}end{align}
        and,
        begin{align}int_0^1 x^{3(n+1)},dx=frac{1}{3n+4}end{align}
        Therefore,
        begin{align}left|int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dxright|leq frac{1}{3n+4}end{align}
        begin{align}left|int_0^1 frac{1}{1+x^3},dx-sum_{k=0}^n frac{(-1)^k}{3k+1}right|leq frac{1}{3n+4}end{align}
        Therefore,
        begin{align}boxed{int_0^1 frac{1}{1+x^3},dx=sum_{k=0}^infty frac{(-1)^k}{3k+1}}end{align}






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          for $x$ real, $ngeq 0$ integer
          begin{align}frac{1}{1+x^3}&=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1-(-x^3)}\
          &=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1+x^3}\
          end{align}



          For $xneq 1$, $ngeq 0$ integer, begin{align}sum_{k=0}^n x^k=frac{1-x^{n+1}}{1-x}end{align}



          Therefore,
          begin{align}int_0^1 frac{1}{1+x^3},dx&=int_0^1 left(sum_{k=0}^n (-x^3)^kright),dx+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
          &=sum_{k=0}^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
          &=sum_{k=0}^n frac{(-1)^k}{3k+1}+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
          end{align}



          For $xin[0;1],ngeq 0$, integer,
          begin{align}frac{x^{3(n+1)}}{1+x^3}leq x^{3(n+1)}end{align}
          and,
          begin{align}int_0^1 x^{3(n+1)},dx=frac{1}{3n+4}end{align}
          Therefore,
          begin{align}left|int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dxright|leq frac{1}{3n+4}end{align}
          begin{align}left|int_0^1 frac{1}{1+x^3},dx-sum_{k=0}^n frac{(-1)^k}{3k+1}right|leq frac{1}{3n+4}end{align}
          Therefore,
          begin{align}boxed{int_0^1 frac{1}{1+x^3},dx=sum_{k=0}^infty frac{(-1)^k}{3k+1}}end{align}






          share|cite|improve this answer









          $endgroup$



          for $x$ real, $ngeq 0$ integer
          begin{align}frac{1}{1+x^3}&=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1-(-x^3)}\
          &=frac{1-(-x^3)^{n+1}}{1-(-x^3)}+frac{(-x^3)^{n+1}}{1+x^3}\
          end{align}



          For $xneq 1$, $ngeq 0$ integer, begin{align}sum_{k=0}^n x^k=frac{1-x^{n+1}}{1-x}end{align}



          Therefore,
          begin{align}int_0^1 frac{1}{1+x^3},dx&=int_0^1 left(sum_{k=0}^n (-x^3)^kright),dx+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
          &=sum_{k=0}^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
          &=sum_{k=0}^n frac{(-1)^k}{3k+1}+int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dx\
          end{align}



          For $xin[0;1],ngeq 0$, integer,
          begin{align}frac{x^{3(n+1)}}{1+x^3}leq x^{3(n+1)}end{align}
          and,
          begin{align}int_0^1 x^{3(n+1)},dx=frac{1}{3n+4}end{align}
          Therefore,
          begin{align}left|int_0^1 frac{(-x^3)^{n+1}}{1+x^3},dxright|leq frac{1}{3n+4}end{align}
          begin{align}left|int_0^1 frac{1}{1+x^3},dx-sum_{k=0}^n frac{(-1)^k}{3k+1}right|leq frac{1}{3n+4}end{align}
          Therefore,
          begin{align}boxed{int_0^1 frac{1}{1+x^3},dx=sum_{k=0}^infty frac{(-1)^k}{3k+1}}end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          FDPFDP

          6,12211929




          6,12211929























              1












              $begingroup$

              If $lvert xrvert<1$, let$$f(x)=sum_{n=0}^inftyfrac{x^{3n+1}}{3n+1}.$$Then $$f'(x)=sum_{n=0}^infty x^{3n}=frac1{1-x^3}.$$Thereforebegin{align}1-frac14+frac17-frac1{10}+cdots&=lim_{xto1}f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac1{1-x^3},mathrm dx.end{align}






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If $lvert xrvert<1$, let$$f(x)=sum_{n=0}^inftyfrac{x^{3n+1}}{3n+1}.$$Then $$f'(x)=sum_{n=0}^infty x^{3n}=frac1{1-x^3}.$$Thereforebegin{align}1-frac14+frac17-frac1{10}+cdots&=lim_{xto1}f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac1{1-x^3},mathrm dx.end{align}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If $lvert xrvert<1$, let$$f(x)=sum_{n=0}^inftyfrac{x^{3n+1}}{3n+1}.$$Then $$f'(x)=sum_{n=0}^infty x^{3n}=frac1{1-x^3}.$$Thereforebegin{align}1-frac14+frac17-frac1{10}+cdots&=lim_{xto1}f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac1{1-x^3},mathrm dx.end{align}






                  share|cite|improve this answer









                  $endgroup$



                  If $lvert xrvert<1$, let$$f(x)=sum_{n=0}^inftyfrac{x^{3n+1}}{3n+1}.$$Then $$f'(x)=sum_{n=0}^infty x^{3n}=frac1{1-x^3}.$$Thereforebegin{align}1-frac14+frac17-frac1{10}+cdots&=lim_{xto1}f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac1{1-x^3},mathrm dx.end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  José Carlos SantosJosé Carlos Santos

                  174k23133243




                  174k23133243






























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                      Tabula Rosettana

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                      Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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                      How to label and detect the document text images

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                      1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
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