A Series for $pi$












7












$begingroup$



Question: Can we show that $$sum_{n=0}^infty(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}=frac{pi}{8} $$ ?





According to wolfram alpha this result is true.



Just amateur curiosity, not sure of a starting point to show if this true.






calculus sequences-and-series pi







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    For my own amusement, I will choose to believe this user is actually Sarah Silverman, the actress and comedian.
    $endgroup$
    – The Count
    Dec 30 '18 at 23:36
















7












$begingroup$



Question: Can we show that $$sum_{n=0}^infty(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}=frac{pi}{8} $$ ?





According to wolfram alpha this result is true.



Just amateur curiosity, not sure of a starting point to show if this true.






calculus sequences-and-series pi







share|cite|improve this question









$endgroup$












  • $begingroup$
    For my own amusement, I will choose to believe this user is actually Sarah Silverman, the actress and comedian.
    $endgroup$
    – The Count
    Dec 30 '18 at 23:36














7












7








7


4



$begingroup$



Question: Can we show that $$sum_{n=0}^infty(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}=frac{pi}{8} $$ ?





According to wolfram alpha this result is true.



Just amateur curiosity, not sure of a starting point to show if this true.






calculus sequences-and-series pi







share|cite|improve this question









$endgroup$





Question: Can we show that $$sum_{n=0}^infty(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}=frac{pi}{8} $$ ?





According to wolfram alpha this result is true.



Just amateur curiosity, not sure of a starting point to show if this true.






calculus sequences-and-series pi




calculus sequences-and-series pi






share|cite|improve this question













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share|cite|improve this question




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asked Dec 30 '18 at 23:18









S. SilvermanS. Silverman

497




497












  • $begingroup$
    For my own amusement, I will choose to believe this user is actually Sarah Silverman, the actress and comedian.
    $endgroup$
    – The Count
    Dec 30 '18 at 23:36


















  • $begingroup$
    For my own amusement, I will choose to believe this user is actually Sarah Silverman, the actress and comedian.
    $endgroup$
    – The Count
    Dec 30 '18 at 23:36
















$begingroup$
For my own amusement, I will choose to believe this user is actually Sarah Silverman, the actress and comedian.
$endgroup$
– The Count
Dec 30 '18 at 23:36




$begingroup$
For my own amusement, I will choose to believe this user is actually Sarah Silverman, the actress and comedian.
$endgroup$
– The Count
Dec 30 '18 at 23:36










2 Answers
2






active

oldest

votes


















16












$begingroup$

Note that after cancellation we have



$$frac{(2n-3)!!}{(2n+3)!!}=frac{1}{(2n-1)(2n+1)(2n+3)}$$



Taking partial fractions gives
$$
frac{1}{(2n-1)(2n+1)(2n+3)}=frac{1}{8}left(-frac{1}{2n-1}+frac{2}{2n+1}-frac{1}{2n+3}right)
$$



Now, the standard arctangent series for $pi$ gives $sum_{n=0}^infty (-1)^n frac{1}{2n+1}=frac{pi}{4}$. So
$$sum_{n=0}^infty (-1)^n frac{1}{2n-1}=-sum_{n=-1}^infty (-1)^n frac{1}{2n+1}=-1-frac{pi}{4}$$
$$
sum_{n=0}^infty (-1)^n frac{1}{2n-3}=-sum_{n=1}^infty (-1)^n frac{1}{2n+1}=1-frac{pi}{4}
$$
which means that when we combine all three infinite sums we get
$$
frac{1}{8}left[-left(-1-frac{pi}{4}right)+2left(frac{pi}{4}right)-left(1-frac{pi}{4}right)right]=frac{pi}{8}
$$
as desired.



(This solution makes it obvious that the sum should be of the form $a+bpi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $pi$.)






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.



    Let
    $$S_p(x)=sum_{n=0}^p(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write
    $$S_p(x)=frac{left(8 p^3+36 p^2+46 p+15right) ,
    _2F_1left(-frac{1}{2},1;frac{5}{2};-xright)+3 (-x)^{p+1} ,
    _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-xright)}{3 left(8 p^3+36 p^2+46
    p+15right)}$$     Using
    $$,
    _2F_1left(-frac{1}{2},1;frac{5}{2};-1right)=frac{3 pi }{8}$$     then
    $$S_p(1)=frac{pi left(8 p^3+36 p^2+46 p+15right)-8 (-1)^p ,
    _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)}{8 left(8 p^3+36 p^2+46
    p+15right)}$$     that it to say
    $$S_p(1)=frac pi 8-frac{(-1)^p }{8 p^3+36
    p^2+46 p+15}, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$$     and $, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$ looks like an hyperbolic function going asymptotically to $frac 12$.



    Then $S_infty(1)=frac{ pi }{8}$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      active

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      16












      $begingroup$

      Note that after cancellation we have



      $$frac{(2n-3)!!}{(2n+3)!!}=frac{1}{(2n-1)(2n+1)(2n+3)}$$



      Taking partial fractions gives
      $$
      frac{1}{(2n-1)(2n+1)(2n+3)}=frac{1}{8}left(-frac{1}{2n-1}+frac{2}{2n+1}-frac{1}{2n+3}right)
      $$



      Now, the standard arctangent series for $pi$ gives $sum_{n=0}^infty (-1)^n frac{1}{2n+1}=frac{pi}{4}$. So
      $$sum_{n=0}^infty (-1)^n frac{1}{2n-1}=-sum_{n=-1}^infty (-1)^n frac{1}{2n+1}=-1-frac{pi}{4}$$
      $$
      sum_{n=0}^infty (-1)^n frac{1}{2n-3}=-sum_{n=1}^infty (-1)^n frac{1}{2n+1}=1-frac{pi}{4}
      $$
      which means that when we combine all three infinite sums we get
      $$
      frac{1}{8}left[-left(-1-frac{pi}{4}right)+2left(frac{pi}{4}right)-left(1-frac{pi}{4}right)right]=frac{pi}{8}
      $$
      as desired.



      (This solution makes it obvious that the sum should be of the form $a+bpi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $pi$.)






      share|cite|improve this answer











      $endgroup$


















        16












        $begingroup$

        Note that after cancellation we have



        $$frac{(2n-3)!!}{(2n+3)!!}=frac{1}{(2n-1)(2n+1)(2n+3)}$$



        Taking partial fractions gives
        $$
        frac{1}{(2n-1)(2n+1)(2n+3)}=frac{1}{8}left(-frac{1}{2n-1}+frac{2}{2n+1}-frac{1}{2n+3}right)
        $$



        Now, the standard arctangent series for $pi$ gives $sum_{n=0}^infty (-1)^n frac{1}{2n+1}=frac{pi}{4}$. So
        $$sum_{n=0}^infty (-1)^n frac{1}{2n-1}=-sum_{n=-1}^infty (-1)^n frac{1}{2n+1}=-1-frac{pi}{4}$$
        $$
        sum_{n=0}^infty (-1)^n frac{1}{2n-3}=-sum_{n=1}^infty (-1)^n frac{1}{2n+1}=1-frac{pi}{4}
        $$
        which means that when we combine all three infinite sums we get
        $$
        frac{1}{8}left[-left(-1-frac{pi}{4}right)+2left(frac{pi}{4}right)-left(1-frac{pi}{4}right)right]=frac{pi}{8}
        $$
        as desired.



        (This solution makes it obvious that the sum should be of the form $a+bpi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $pi$.)






        share|cite|improve this answer











        $endgroup$
















          16












          16








          16





          $begingroup$

          Note that after cancellation we have



          $$frac{(2n-3)!!}{(2n+3)!!}=frac{1}{(2n-1)(2n+1)(2n+3)}$$



          Taking partial fractions gives
          $$
          frac{1}{(2n-1)(2n+1)(2n+3)}=frac{1}{8}left(-frac{1}{2n-1}+frac{2}{2n+1}-frac{1}{2n+3}right)
          $$



          Now, the standard arctangent series for $pi$ gives $sum_{n=0}^infty (-1)^n frac{1}{2n+1}=frac{pi}{4}$. So
          $$sum_{n=0}^infty (-1)^n frac{1}{2n-1}=-sum_{n=-1}^infty (-1)^n frac{1}{2n+1}=-1-frac{pi}{4}$$
          $$
          sum_{n=0}^infty (-1)^n frac{1}{2n-3}=-sum_{n=1}^infty (-1)^n frac{1}{2n+1}=1-frac{pi}{4}
          $$
          which means that when we combine all three infinite sums we get
          $$
          frac{1}{8}left[-left(-1-frac{pi}{4}right)+2left(frac{pi}{4}right)-left(1-frac{pi}{4}right)right]=frac{pi}{8}
          $$
          as desired.



          (This solution makes it obvious that the sum should be of the form $a+bpi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $pi$.)






          share|cite|improve this answer











          $endgroup$



          Note that after cancellation we have



          $$frac{(2n-3)!!}{(2n+3)!!}=frac{1}{(2n-1)(2n+1)(2n+3)}$$



          Taking partial fractions gives
          $$
          frac{1}{(2n-1)(2n+1)(2n+3)}=frac{1}{8}left(-frac{1}{2n-1}+frac{2}{2n+1}-frac{1}{2n+3}right)
          $$



          Now, the standard arctangent series for $pi$ gives $sum_{n=0}^infty (-1)^n frac{1}{2n+1}=frac{pi}{4}$. So
          $$sum_{n=0}^infty (-1)^n frac{1}{2n-1}=-sum_{n=-1}^infty (-1)^n frac{1}{2n+1}=-1-frac{pi}{4}$$
          $$
          sum_{n=0}^infty (-1)^n frac{1}{2n-3}=-sum_{n=1}^infty (-1)^n frac{1}{2n+1}=1-frac{pi}{4}
          $$
          which means that when we combine all three infinite sums we get
          $$
          frac{1}{8}left[-left(-1-frac{pi}{4}right)+2left(frac{pi}{4}right)-left(1-frac{pi}{4}right)right]=frac{pi}{8}
          $$
          as desired.



          (This solution makes it obvious that the sum should be of the form $a+bpi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $pi$.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 23:37

























          answered Dec 30 '18 at 23:31









          MicahMicah

          30.2k1364106




          30.2k1364106























              3












              $begingroup$

              Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.



              Let
              $$S_p(x)=sum_{n=0}^p(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write
              $$S_p(x)=frac{left(8 p^3+36 p^2+46 p+15right) ,
              _2F_1left(-frac{1}{2},1;frac{5}{2};-xright)+3 (-x)^{p+1} ,
              _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-xright)}{3 left(8 p^3+36 p^2+46
              p+15right)}$$               Using
              $$,
              _2F_1left(-frac{1}{2},1;frac{5}{2};-1right)=frac{3 pi }{8}$$               then
              $$S_p(1)=frac{pi left(8 p^3+36 p^2+46 p+15right)-8 (-1)^p ,
              _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)}{8 left(8 p^3+36 p^2+46
              p+15right)}$$               that it to say
              $$S_p(1)=frac pi 8-frac{(-1)^p }{8 p^3+36
              p^2+46 p+15}, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$$               and $, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$ looks like an hyperbolic function going asymptotically to $frac 12$.



              Then $S_infty(1)=frac{ pi }{8}$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.



                Let
                $$S_p(x)=sum_{n=0}^p(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write
                $$S_p(x)=frac{left(8 p^3+36 p^2+46 p+15right) ,
                _2F_1left(-frac{1}{2},1;frac{5}{2};-xright)+3 (-x)^{p+1} ,
                _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-xright)}{3 left(8 p^3+36 p^2+46
                p+15right)}$$                 Using
                $$,
                _2F_1left(-frac{1}{2},1;frac{5}{2};-1right)=frac{3 pi }{8}$$                 then
                $$S_p(1)=frac{pi left(8 p^3+36 p^2+46 p+15right)-8 (-1)^p ,
                _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)}{8 left(8 p^3+36 p^2+46
                p+15right)}$$                 that it to say
                $$S_p(1)=frac pi 8-frac{(-1)^p }{8 p^3+36
                p^2+46 p+15}, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$$                 and $, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$ looks like an hyperbolic function going asymptotically to $frac 12$.



                Then $S_infty(1)=frac{ pi }{8}$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.



                  Let
                  $$S_p(x)=sum_{n=0}^p(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write
                  $$S_p(x)=frac{left(8 p^3+36 p^2+46 p+15right) ,
                  _2F_1left(-frac{1}{2},1;frac{5}{2};-xright)+3 (-x)^{p+1} ,
                  _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-xright)}{3 left(8 p^3+36 p^2+46
                  p+15right)}$$                   Using
                  $$,
                  _2F_1left(-frac{1}{2},1;frac{5}{2};-1right)=frac{3 pi }{8}$$                   then
                  $$S_p(1)=frac{pi left(8 p^3+36 p^2+46 p+15right)-8 (-1)^p ,
                  _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)}{8 left(8 p^3+36 p^2+46
                  p+15right)}$$                   that it to say
                  $$S_p(1)=frac pi 8-frac{(-1)^p }{8 p^3+36
                  p^2+46 p+15}, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$$                   and $, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$ looks like an hyperbolic function going asymptotically to $frac 12$.



                  Then $S_infty(1)=frac{ pi }{8}$.






                  share|cite|improve this answer









                  $endgroup$



                  Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.



                  Let
                  $$S_p(x)=sum_{n=0}^p(-1)^{n+1}frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write
                  $$S_p(x)=frac{left(8 p^3+36 p^2+46 p+15right) ,
                  _2F_1left(-frac{1}{2},1;frac{5}{2};-xright)+3 (-x)^{p+1} ,
                  _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-xright)}{3 left(8 p^3+36 p^2+46
                  p+15right)}$$                   Using
                  $$,
                  _2F_1left(-frac{1}{2},1;frac{5}{2};-1right)=frac{3 pi }{8}$$                   then
                  $$S_p(1)=frac{pi left(8 p^3+36 p^2+46 p+15right)-8 (-1)^p ,
                  _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)}{8 left(8 p^3+36 p^2+46
                  p+15right)}$$                   that it to say
                  $$S_p(1)=frac pi 8-frac{(-1)^p }{8 p^3+36
                  p^2+46 p+15}, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$$                   and $, _2F_1left(1,p+frac{1}{2};p+frac{7}{2};-1right)$ looks like an hyperbolic function going asymptotically to $frac 12$.



                  Then $S_infty(1)=frac{ pi }{8}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 4:34









                  Claude LeiboviciClaude Leibovici

                  123k1157134




                  123k1157134






























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