Determining the infinite sum $sumlimits_{n=0}^{infty}frac{(-1)^n}{2n+1}$












0












$begingroup$


I have to determine the following sum:



begin{equation}
sumlimits_{n=0}^{infty}frac{(-1)^n}{2n+1}
end{equation}



And i've tried lot's of ways to determine it, but i couldn't get a result so I figured I would look on WolframAlpha to see what the result has to be so i could maybe know what to do next by that. But apparently the result is $frac{pi}{4}$ and i don't know how I could possibly get to pi.
Does anyone have any suggestions about how to tackle this?






sequences-and-series summation







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  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 18:23
















0












$begingroup$


I have to determine the following sum:



begin{equation}
sumlimits_{n=0}^{infty}frac{(-1)^n}{2n+1}
end{equation}



And i've tried lot's of ways to determine it, but i couldn't get a result so I figured I would look on WolframAlpha to see what the result has to be so i could maybe know what to do next by that. But apparently the result is $frac{pi}{4}$ and i don't know how I could possibly get to pi.
Does anyone have any suggestions about how to tackle this?






sequences-and-series summation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 18:23














0












0








0





$begingroup$


I have to determine the following sum:



begin{equation}
sumlimits_{n=0}^{infty}frac{(-1)^n}{2n+1}
end{equation}



And i've tried lot's of ways to determine it, but i couldn't get a result so I figured I would look on WolframAlpha to see what the result has to be so i could maybe know what to do next by that. But apparently the result is $frac{pi}{4}$ and i don't know how I could possibly get to pi.
Does anyone have any suggestions about how to tackle this?






sequences-and-series summation







share|cite|improve this question









$endgroup$




I have to determine the following sum:



begin{equation}
sumlimits_{n=0}^{infty}frac{(-1)^n}{2n+1}
end{equation}



And i've tried lot's of ways to determine it, but i couldn't get a result so I figured I would look on WolframAlpha to see what the result has to be so i could maybe know what to do next by that. But apparently the result is $frac{pi}{4}$ and i don't know how I could possibly get to pi.
Does anyone have any suggestions about how to tackle this?






sequences-and-series summation




sequences-and-series summation






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asked Jan 7 at 18:22









ViktorViktor

1389




1389








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 18:23














  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 18:23








1




1




$begingroup$
en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:23




$begingroup$
en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
$endgroup$
– Lord Shark the Unknown
Jan 7 at 18:23










4 Answers
4






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5












$begingroup$

Observe $$frac{1}{1-x}=sum_{n=0}^infty x^n$$



$$frac{1}{1+x^2}=sum_{n=0}^infty (-1)^nx^{2n}$$



Integrating both sides from $0$ to $1$ we get $$arctan(x)|_0^1=sum_{n=0}^infty frac{(-1)^n x^{2n+1}}{2n+1}|_0^1$$
so $$arctan(1)=frac{pi}{4}=sum_{n=0}^infty frac{(-1)^n}{2n+1}$$






share|cite|improve this answer









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  • $begingroup$
    Thanks i now understand why this is the case!
    $endgroup$
    – Viktor
    Jan 7 at 18:30



















1












$begingroup$

Hint: $sum_{k=0}^n{frac{(-1)^k}{2k+1}} = int_0^1{frac{1}{1+x^2}} - int_0^1{frac{(-x^2)^{n+1}}{1+x^2}}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$


    $$frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}$$ This is because, let $y=tan^{-1}(x)$, so that $$x=tan(y)tag{1}$$ Differentiating both sides of $(1)$ yields $$1=frac{1}{cos^2(y)}frac{dy}{dx}\ 1=frac{cos^2(y)+sin^2(y)}{cos^2(y)}frac{dy}{dx}\\1=left(1+x^2right)frac{dy}{dx}$$ $$boxed{frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}}tag{2}$$




    Now, it is known that $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$ for $|x|<1$, replacing $x$ with $-x^2$ we get $$frac{1}{1+x^2}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{3}$$ Combining $(2)$ and $(3)$ we see that $$frac{d(tan^{-1}(x))}{dx}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{4}$$ Taking the integral from $0$ to $1$ of $(4)$ yields $$int_0^1frac{d(tan^{-1}(x))}{dx}=tan^{-1}(1)-tan^{-1}(0)=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}\boxed{frac{pi}{4}=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}}$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      As others have shown,
      $$sum_{ngeq0}frac{(-1)^n}{2n+1}=int_0^1frac{mathrm dx}{1+x^2}$$
      but we can generalize this.



      $$frac1{1+x}=sum_{ngeq0}(-1)^nx^n$$
      $$frac1{1+x^a}=sum_{ngeq0}(-1)^nx^{an}$$
      $$int_0^1frac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^n}{an+1}$$
      Also
      $$int_0^bfrac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^nb^{an+1}}{an+1}$$






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

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        active

        oldest

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        active

        oldest

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        5












        $begingroup$

        Observe $$frac{1}{1-x}=sum_{n=0}^infty x^n$$



        $$frac{1}{1+x^2}=sum_{n=0}^infty (-1)^nx^{2n}$$



        Integrating both sides from $0$ to $1$ we get $$arctan(x)|_0^1=sum_{n=0}^infty frac{(-1)^n x^{2n+1}}{2n+1}|_0^1$$
        so $$arctan(1)=frac{pi}{4}=sum_{n=0}^infty frac{(-1)^n}{2n+1}$$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thanks i now understand why this is the case!
          $endgroup$
          – Viktor
          Jan 7 at 18:30
















        5












        $begingroup$

        Observe $$frac{1}{1-x}=sum_{n=0}^infty x^n$$



        $$frac{1}{1+x^2}=sum_{n=0}^infty (-1)^nx^{2n}$$



        Integrating both sides from $0$ to $1$ we get $$arctan(x)|_0^1=sum_{n=0}^infty frac{(-1)^n x^{2n+1}}{2n+1}|_0^1$$
        so $$arctan(1)=frac{pi}{4}=sum_{n=0}^infty frac{(-1)^n}{2n+1}$$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Thanks i now understand why this is the case!
          $endgroup$
          – Viktor
          Jan 7 at 18:30














        5












        5








        5





        $begingroup$

        Observe $$frac{1}{1-x}=sum_{n=0}^infty x^n$$



        $$frac{1}{1+x^2}=sum_{n=0}^infty (-1)^nx^{2n}$$



        Integrating both sides from $0$ to $1$ we get $$arctan(x)|_0^1=sum_{n=0}^infty frac{(-1)^n x^{2n+1}}{2n+1}|_0^1$$
        so $$arctan(1)=frac{pi}{4}=sum_{n=0}^infty frac{(-1)^n}{2n+1}$$






        share|cite|improve this answer









        $endgroup$



        Observe $$frac{1}{1-x}=sum_{n=0}^infty x^n$$



        $$frac{1}{1+x^2}=sum_{n=0}^infty (-1)^nx^{2n}$$



        Integrating both sides from $0$ to $1$ we get $$arctan(x)|_0^1=sum_{n=0}^infty frac{(-1)^n x^{2n+1}}{2n+1}|_0^1$$
        so $$arctan(1)=frac{pi}{4}=sum_{n=0}^infty frac{(-1)^n}{2n+1}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 18:28









        aledenaleden

        2,032511




        2,032511












        • $begingroup$
          Thanks i now understand why this is the case!
          $endgroup$
          – Viktor
          Jan 7 at 18:30


















        • $begingroup$
          Thanks i now understand why this is the case!
          $endgroup$
          – Viktor
          Jan 7 at 18:30
















        $begingroup$
        Thanks i now understand why this is the case!
        $endgroup$
        – Viktor
        Jan 7 at 18:30




        $begingroup$
        Thanks i now understand why this is the case!
        $endgroup$
        – Viktor
        Jan 7 at 18:30











        1












        $begingroup$

        Hint: $sum_{k=0}^n{frac{(-1)^k}{2k+1}} = int_0^1{frac{1}{1+x^2}} - int_0^1{frac{(-x^2)^{n+1}}{1+x^2}}$.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Hint: $sum_{k=0}^n{frac{(-1)^k}{2k+1}} = int_0^1{frac{1}{1+x^2}} - int_0^1{frac{(-x^2)^{n+1}}{1+x^2}}$.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint: $sum_{k=0}^n{frac{(-1)^k}{2k+1}} = int_0^1{frac{1}{1+x^2}} - int_0^1{frac{(-x^2)^{n+1}}{1+x^2}}$.






            share|cite|improve this answer









            $endgroup$



            Hint: $sum_{k=0}^n{frac{(-1)^k}{2k+1}} = int_0^1{frac{1}{1+x^2}} - int_0^1{frac{(-x^2)^{n+1}}{1+x^2}}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 18:30









            MindlackMindlack

            3,37717




            3,37717























                1












                $begingroup$


                $$frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}$$ This is because, let $y=tan^{-1}(x)$, so that $$x=tan(y)tag{1}$$ Differentiating both sides of $(1)$ yields $$1=frac{1}{cos^2(y)}frac{dy}{dx}\ 1=frac{cos^2(y)+sin^2(y)}{cos^2(y)}frac{dy}{dx}\\1=left(1+x^2right)frac{dy}{dx}$$ $$boxed{frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}}tag{2}$$




                Now, it is known that $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$ for $|x|<1$, replacing $x$ with $-x^2$ we get $$frac{1}{1+x^2}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{3}$$ Combining $(2)$ and $(3)$ we see that $$frac{d(tan^{-1}(x))}{dx}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{4}$$ Taking the integral from $0$ to $1$ of $(4)$ yields $$int_0^1frac{d(tan^{-1}(x))}{dx}=tan^{-1}(1)-tan^{-1}(0)=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}\boxed{frac{pi}{4}=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}}$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$


                  $$frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}$$ This is because, let $y=tan^{-1}(x)$, so that $$x=tan(y)tag{1}$$ Differentiating both sides of $(1)$ yields $$1=frac{1}{cos^2(y)}frac{dy}{dx}\ 1=frac{cos^2(y)+sin^2(y)}{cos^2(y)}frac{dy}{dx}\\1=left(1+x^2right)frac{dy}{dx}$$ $$boxed{frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}}tag{2}$$




                  Now, it is known that $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$ for $|x|<1$, replacing $x$ with $-x^2$ we get $$frac{1}{1+x^2}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{3}$$ Combining $(2)$ and $(3)$ we see that $$frac{d(tan^{-1}(x))}{dx}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{4}$$ Taking the integral from $0$ to $1$ of $(4)$ yields $$int_0^1frac{d(tan^{-1}(x))}{dx}=tan^{-1}(1)-tan^{-1}(0)=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}\boxed{frac{pi}{4}=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}}$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$


                    $$frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}$$ This is because, let $y=tan^{-1}(x)$, so that $$x=tan(y)tag{1}$$ Differentiating both sides of $(1)$ yields $$1=frac{1}{cos^2(y)}frac{dy}{dx}\ 1=frac{cos^2(y)+sin^2(y)}{cos^2(y)}frac{dy}{dx}\\1=left(1+x^2right)frac{dy}{dx}$$ $$boxed{frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}}tag{2}$$




                    Now, it is known that $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$ for $|x|<1$, replacing $x$ with $-x^2$ we get $$frac{1}{1+x^2}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{3}$$ Combining $(2)$ and $(3)$ we see that $$frac{d(tan^{-1}(x))}{dx}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{4}$$ Taking the integral from $0$ to $1$ of $(4)$ yields $$int_0^1frac{d(tan^{-1}(x))}{dx}=tan^{-1}(1)-tan^{-1}(0)=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}\boxed{frac{pi}{4}=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}}$$






                    share|cite|improve this answer









                    $endgroup$




                    $$frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}$$ This is because, let $y=tan^{-1}(x)$, so that $$x=tan(y)tag{1}$$ Differentiating both sides of $(1)$ yields $$1=frac{1}{cos^2(y)}frac{dy}{dx}\ 1=frac{cos^2(y)+sin^2(y)}{cos^2(y)}frac{dy}{dx}\\1=left(1+x^2right)frac{dy}{dx}$$ $$boxed{frac{d(tan^{-1}(x))}{dx}=frac{1}{1+x^2}}tag{2}$$




                    Now, it is known that $$frac{1}{1-x}=sum_{n=0}^{infty}x^n$$ for $|x|<1$, replacing $x$ with $-x^2$ we get $$frac{1}{1+x^2}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{3}$$ Combining $(2)$ and $(3)$ we see that $$frac{d(tan^{-1}(x))}{dx}=sum_{n=0}^{infty}(-1)^nx^{2n}tag{4}$$ Taking the integral from $0$ to $1$ of $(4)$ yields $$int_0^1frac{d(tan^{-1}(x))}{dx}=tan^{-1}(1)-tan^{-1}(0)=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}\boxed{frac{pi}{4}=sum_{n=0}^{infty}frac{(-1)^n}{2n+1}}$$







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                    answered Jan 7 at 18:49









                    cansomeonehelpmeoutcansomeonehelpmeout

                    6,8273835




                    6,8273835























                        1












                        $begingroup$

                        As others have shown,
                        $$sum_{ngeq0}frac{(-1)^n}{2n+1}=int_0^1frac{mathrm dx}{1+x^2}$$
                        but we can generalize this.



                        $$frac1{1+x}=sum_{ngeq0}(-1)^nx^n$$
                        $$frac1{1+x^a}=sum_{ngeq0}(-1)^nx^{an}$$
                        $$int_0^1frac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^n}{an+1}$$
                        Also
                        $$int_0^bfrac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^nb^{an+1}}{an+1}$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          As others have shown,
                          $$sum_{ngeq0}frac{(-1)^n}{2n+1}=int_0^1frac{mathrm dx}{1+x^2}$$
                          but we can generalize this.



                          $$frac1{1+x}=sum_{ngeq0}(-1)^nx^n$$
                          $$frac1{1+x^a}=sum_{ngeq0}(-1)^nx^{an}$$
                          $$int_0^1frac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^n}{an+1}$$
                          Also
                          $$int_0^bfrac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^nb^{an+1}}{an+1}$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            As others have shown,
                            $$sum_{ngeq0}frac{(-1)^n}{2n+1}=int_0^1frac{mathrm dx}{1+x^2}$$
                            but we can generalize this.



                            $$frac1{1+x}=sum_{ngeq0}(-1)^nx^n$$
                            $$frac1{1+x^a}=sum_{ngeq0}(-1)^nx^{an}$$
                            $$int_0^1frac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^n}{an+1}$$
                            Also
                            $$int_0^bfrac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^nb^{an+1}}{an+1}$$






                            share|cite|improve this answer









                            $endgroup$



                            As others have shown,
                            $$sum_{ngeq0}frac{(-1)^n}{2n+1}=int_0^1frac{mathrm dx}{1+x^2}$$
                            but we can generalize this.



                            $$frac1{1+x}=sum_{ngeq0}(-1)^nx^n$$
                            $$frac1{1+x^a}=sum_{ngeq0}(-1)^nx^{an}$$
                            $$int_0^1frac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^n}{an+1}$$
                            Also
                            $$int_0^bfrac{mathrm dx}{1+x^a}=sum_{ngeq0}frac{(-1)^nb^{an+1}}{an+1}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 7 at 19:04









                            clathratusclathratus

                            3,952334




                            3,952334






























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