Proof of this formula for $sqrt{epi/2}$ and similar formulas.












4












$begingroup$


begin{align}
sqrt{frac{epi}{2}}=1+frac{1}{1cdot3}+frac{1}{1cdot3cdot5}+frac{1}{1cdot3cdot5cdot7}+dots+cfrac1{1+cfrac{1}{1+cfrac{2}{1+cfrac{3}{1+ddots}}}}
end{align}



as seen here.



Is there other series that relate $pi$ and $e$?



Also, it's possible to rewrite the continued fraction above in terms of known functions/numbers?






sequences-and-series exponential-function pi continued-fractions







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    begin{align}
    sqrt{frac{epi}{2}}=1+frac{1}{1cdot3}+frac{1}{1cdot3cdot5}+frac{1}{1cdot3cdot5cdot7}+dots+cfrac1{1+cfrac{1}{1+cfrac{2}{1+cfrac{3}{1+ddots}}}}
    end{align}



    as seen here.



    Is there other series that relate $pi$ and $e$?



    Also, it's possible to rewrite the continued fraction above in terms of known functions/numbers?






    sequences-and-series exponential-function pi continued-fractions







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      3



      $begingroup$


      begin{align}
      sqrt{frac{epi}{2}}=1+frac{1}{1cdot3}+frac{1}{1cdot3cdot5}+frac{1}{1cdot3cdot5cdot7}+dots+cfrac1{1+cfrac{1}{1+cfrac{2}{1+cfrac{3}{1+ddots}}}}
      end{align}



      as seen here.



      Is there other series that relate $pi$ and $e$?



      Also, it's possible to rewrite the continued fraction above in terms of known functions/numbers?






      sequences-and-series exponential-function pi continued-fractions







      share|cite|improve this question











      $endgroup$




      begin{align}
      sqrt{frac{epi}{2}}=1+frac{1}{1cdot3}+frac{1}{1cdot3cdot5}+frac{1}{1cdot3cdot5cdot7}+dots+cfrac1{1+cfrac{1}{1+cfrac{2}{1+cfrac{3}{1+ddots}}}}
      end{align}



      as seen here.



      Is there other series that relate $pi$ and $e$?



      Also, it's possible to rewrite the continued fraction above in terms of known functions/numbers?






      sequences-and-series exponential-function pi continued-fractions




      sequences-and-series exponential-function pi continued-fractions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 6:41









      Lee David Chung Lin

      4,39341242




      4,39341242










      asked Jan 6 at 4:20









      PintecoPinteco

      765313




      765313






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          The infinite sum is
          $,sqrt{e pi/2},textrm{erf}(sqrt{1/2}),$
          as given in OEIS sequence A060196. The
          continued fraction is $,sqrt{e pi/2},textrm{erfc}(sqrt{1/2}),$
          as given in OIES sequence A108088. The sum, of course, is $,sqrt{e pi/2},$ since
          $,textrm{erf}(x) + textrm{erfc}(x) = 1,$ by definition.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            About 2 years ago I discovered a lot of pretty nice series that relate $pi$ and $e$, for instance :
            $$sum_{n=1}^{infty}frac{n^2}{16n^4-1}=frac{pi}{32}cdotfrac{e^{pi}+1}{e^{pi}-1}$$
            $$sum_{n=1}^{infty}frac{n^2}{4n^4+1}=frac{pi}{8}cdotfrac{e^{pi}-1}{e^{pi}+1}$$
            $$sum_{n=1}^{infty}frac{n^2}{(4n^4+1)(16n^4-1)}=frac{pi}{10}cdotfrac{e^{pi}}{e^{2pi}-1}$$
            $$sum_{n=1}^{infty}frac{n^2(32n^4+3)}{(4n^4+1)(16n^4-1)}=frac{pi}{4}cdotfrac{e^{2pi}+1}{e^{2pi}-1}$$
            $$sum_{n=1}^{infty}frac{n^2(64n^4+11)}{(4n^4+1)(16n^4-1)}=frac{pi}{2}cdotfrac{e^{3pi}-1}{(e^{2pi}-1)(e^{pi}-1)}$$



            If you're looking for any mathematical identity that relates $pi$ and $e$, I can also suggest :



            $cdot$ The Stirling limit : $$lim_{ntoinfty}frac{n!e^n}{n^nsqrt{n}}=sqrt{2pi}$$



            $cdot$ The well known integral : $$int_{-infty}^{infty}frac{cos(x)}{x^2+1}text{d}x=frac{pi}{e}$$



            $cdot$ Victor Adamchik's integrals : $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x+1)^2+pi^2}=frac{1}{2}$$ $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x)^2+pi^2}=frac{1}{1+Omega}$$
            Where $Omega$ is the mathematical constant defined by $text{ }Omega e^{Omega}=1$.



            Hope this helps.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Where can I read more about these series and other similar ones? Is there any paper or a link?
              $endgroup$
              – Pinteco
              Jan 6 at 19:45










            • $begingroup$
              Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
              $endgroup$
              – Harmonic Sun
              Jan 6 at 19:56










            • $begingroup$
              @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
              $endgroup$
              – Jack D'Aurizio
              Jan 7 at 0:24











            Your Answer





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            2 Answers
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            2 Answers
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            4












            $begingroup$

            The infinite sum is
            $,sqrt{e pi/2},textrm{erf}(sqrt{1/2}),$
            as given in OEIS sequence A060196. The
            continued fraction is $,sqrt{e pi/2},textrm{erfc}(sqrt{1/2}),$
            as given in OIES sequence A108088. The sum, of course, is $,sqrt{e pi/2},$ since
            $,textrm{erf}(x) + textrm{erfc}(x) = 1,$ by definition.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The infinite sum is
              $,sqrt{e pi/2},textrm{erf}(sqrt{1/2}),$
              as given in OEIS sequence A060196. The
              continued fraction is $,sqrt{e pi/2},textrm{erfc}(sqrt{1/2}),$
              as given in OIES sequence A108088. The sum, of course, is $,sqrt{e pi/2},$ since
              $,textrm{erf}(x) + textrm{erfc}(x) = 1,$ by definition.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The infinite sum is
                $,sqrt{e pi/2},textrm{erf}(sqrt{1/2}),$
                as given in OEIS sequence A060196. The
                continued fraction is $,sqrt{e pi/2},textrm{erfc}(sqrt{1/2}),$
                as given in OIES sequence A108088. The sum, of course, is $,sqrt{e pi/2},$ since
                $,textrm{erf}(x) + textrm{erfc}(x) = 1,$ by definition.






                share|cite|improve this answer









                $endgroup$



                The infinite sum is
                $,sqrt{e pi/2},textrm{erf}(sqrt{1/2}),$
                as given in OEIS sequence A060196. The
                continued fraction is $,sqrt{e pi/2},textrm{erfc}(sqrt{1/2}),$
                as given in OIES sequence A108088. The sum, of course, is $,sqrt{e pi/2},$ since
                $,textrm{erf}(x) + textrm{erfc}(x) = 1,$ by definition.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 5:41









                SomosSomos

                14.6k11336




                14.6k11336























                    2












                    $begingroup$

                    About 2 years ago I discovered a lot of pretty nice series that relate $pi$ and $e$, for instance :
                    $$sum_{n=1}^{infty}frac{n^2}{16n^4-1}=frac{pi}{32}cdotfrac{e^{pi}+1}{e^{pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{4n^4+1}=frac{pi}{8}cdotfrac{e^{pi}-1}{e^{pi}+1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{(4n^4+1)(16n^4-1)}=frac{pi}{10}cdotfrac{e^{pi}}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(32n^4+3)}{(4n^4+1)(16n^4-1)}=frac{pi}{4}cdotfrac{e^{2pi}+1}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(64n^4+11)}{(4n^4+1)(16n^4-1)}=frac{pi}{2}cdotfrac{e^{3pi}-1}{(e^{2pi}-1)(e^{pi}-1)}$$



                    If you're looking for any mathematical identity that relates $pi$ and $e$, I can also suggest :



                    $cdot$ The Stirling limit : $$lim_{ntoinfty}frac{n!e^n}{n^nsqrt{n}}=sqrt{2pi}$$



                    $cdot$ The well known integral : $$int_{-infty}^{infty}frac{cos(x)}{x^2+1}text{d}x=frac{pi}{e}$$



                    $cdot$ Victor Adamchik's integrals : $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x+1)^2+pi^2}=frac{1}{2}$$ $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x)^2+pi^2}=frac{1}{1+Omega}$$
                    Where $Omega$ is the mathematical constant defined by $text{ }Omega e^{Omega}=1$.



                    Hope this helps.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Where can I read more about these series and other similar ones? Is there any paper or a link?
                      $endgroup$
                      – Pinteco
                      Jan 6 at 19:45










                    • $begingroup$
                      Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
                      $endgroup$
                      – Harmonic Sun
                      Jan 6 at 19:56










                    • $begingroup$
                      @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
                      $endgroup$
                      – Jack D'Aurizio
                      Jan 7 at 0:24
















                    2












                    $begingroup$

                    About 2 years ago I discovered a lot of pretty nice series that relate $pi$ and $e$, for instance :
                    $$sum_{n=1}^{infty}frac{n^2}{16n^4-1}=frac{pi}{32}cdotfrac{e^{pi}+1}{e^{pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{4n^4+1}=frac{pi}{8}cdotfrac{e^{pi}-1}{e^{pi}+1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{(4n^4+1)(16n^4-1)}=frac{pi}{10}cdotfrac{e^{pi}}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(32n^4+3)}{(4n^4+1)(16n^4-1)}=frac{pi}{4}cdotfrac{e^{2pi}+1}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(64n^4+11)}{(4n^4+1)(16n^4-1)}=frac{pi}{2}cdotfrac{e^{3pi}-1}{(e^{2pi}-1)(e^{pi}-1)}$$



                    If you're looking for any mathematical identity that relates $pi$ and $e$, I can also suggest :



                    $cdot$ The Stirling limit : $$lim_{ntoinfty}frac{n!e^n}{n^nsqrt{n}}=sqrt{2pi}$$



                    $cdot$ The well known integral : $$int_{-infty}^{infty}frac{cos(x)}{x^2+1}text{d}x=frac{pi}{e}$$



                    $cdot$ Victor Adamchik's integrals : $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x+1)^2+pi^2}=frac{1}{2}$$ $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x)^2+pi^2}=frac{1}{1+Omega}$$
                    Where $Omega$ is the mathematical constant defined by $text{ }Omega e^{Omega}=1$.



                    Hope this helps.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Where can I read more about these series and other similar ones? Is there any paper or a link?
                      $endgroup$
                      – Pinteco
                      Jan 6 at 19:45










                    • $begingroup$
                      Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
                      $endgroup$
                      – Harmonic Sun
                      Jan 6 at 19:56










                    • $begingroup$
                      @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
                      $endgroup$
                      – Jack D'Aurizio
                      Jan 7 at 0:24














                    2












                    2








                    2





                    $begingroup$

                    About 2 years ago I discovered a lot of pretty nice series that relate $pi$ and $e$, for instance :
                    $$sum_{n=1}^{infty}frac{n^2}{16n^4-1}=frac{pi}{32}cdotfrac{e^{pi}+1}{e^{pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{4n^4+1}=frac{pi}{8}cdotfrac{e^{pi}-1}{e^{pi}+1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{(4n^4+1)(16n^4-1)}=frac{pi}{10}cdotfrac{e^{pi}}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(32n^4+3)}{(4n^4+1)(16n^4-1)}=frac{pi}{4}cdotfrac{e^{2pi}+1}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(64n^4+11)}{(4n^4+1)(16n^4-1)}=frac{pi}{2}cdotfrac{e^{3pi}-1}{(e^{2pi}-1)(e^{pi}-1)}$$



                    If you're looking for any mathematical identity that relates $pi$ and $e$, I can also suggest :



                    $cdot$ The Stirling limit : $$lim_{ntoinfty}frac{n!e^n}{n^nsqrt{n}}=sqrt{2pi}$$



                    $cdot$ The well known integral : $$int_{-infty}^{infty}frac{cos(x)}{x^2+1}text{d}x=frac{pi}{e}$$



                    $cdot$ Victor Adamchik's integrals : $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x+1)^2+pi^2}=frac{1}{2}$$ $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x)^2+pi^2}=frac{1}{1+Omega}$$
                    Where $Omega$ is the mathematical constant defined by $text{ }Omega e^{Omega}=1$.



                    Hope this helps.






                    share|cite|improve this answer











                    $endgroup$



                    About 2 years ago I discovered a lot of pretty nice series that relate $pi$ and $e$, for instance :
                    $$sum_{n=1}^{infty}frac{n^2}{16n^4-1}=frac{pi}{32}cdotfrac{e^{pi}+1}{e^{pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{4n^4+1}=frac{pi}{8}cdotfrac{e^{pi}-1}{e^{pi}+1}$$
                    $$sum_{n=1}^{infty}frac{n^2}{(4n^4+1)(16n^4-1)}=frac{pi}{10}cdotfrac{e^{pi}}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(32n^4+3)}{(4n^4+1)(16n^4-1)}=frac{pi}{4}cdotfrac{e^{2pi}+1}{e^{2pi}-1}$$
                    $$sum_{n=1}^{infty}frac{n^2(64n^4+11)}{(4n^4+1)(16n^4-1)}=frac{pi}{2}cdotfrac{e^{3pi}-1}{(e^{2pi}-1)(e^{pi}-1)}$$



                    If you're looking for any mathematical identity that relates $pi$ and $e$, I can also suggest :



                    $cdot$ The Stirling limit : $$lim_{ntoinfty}frac{n!e^n}{n^nsqrt{n}}=sqrt{2pi}$$



                    $cdot$ The well known integral : $$int_{-infty}^{infty}frac{cos(x)}{x^2+1}text{d}x=frac{pi}{e}$$



                    $cdot$ Victor Adamchik's integrals : $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x+1)^2+pi^2}=frac{1}{2}$$ $$int_{-infty}^{infty}frac{text{d}x}{(e^x-x)^2+pi^2}=frac{1}{1+Omega}$$
                    Where $Omega$ is the mathematical constant defined by $text{ }Omega e^{Omega}=1$.



                    Hope this helps.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 6 at 14:15

























                    answered Jan 6 at 13:59









                    Harmonic SunHarmonic Sun

                    70710




                    70710












                    • $begingroup$
                      Where can I read more about these series and other similar ones? Is there any paper or a link?
                      $endgroup$
                      – Pinteco
                      Jan 6 at 19:45










                    • $begingroup$
                      Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
                      $endgroup$
                      – Harmonic Sun
                      Jan 6 at 19:56










                    • $begingroup$
                      @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
                      $endgroup$
                      – Jack D'Aurizio
                      Jan 7 at 0:24


















                    • $begingroup$
                      Where can I read more about these series and other similar ones? Is there any paper or a link?
                      $endgroup$
                      – Pinteco
                      Jan 6 at 19:45










                    • $begingroup$
                      Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
                      $endgroup$
                      – Harmonic Sun
                      Jan 6 at 19:56










                    • $begingroup$
                      @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
                      $endgroup$
                      – Jack D'Aurizio
                      Jan 7 at 0:24
















                    $begingroup$
                    Where can I read more about these series and other similar ones? Is there any paper or a link?
                    $endgroup$
                    – Pinteco
                    Jan 6 at 19:45




                    $begingroup$
                    Where can I read more about these series and other similar ones? Is there any paper or a link?
                    $endgroup$
                    – Pinteco
                    Jan 6 at 19:45












                    $begingroup$
                    Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
                    $endgroup$
                    – Harmonic Sun
                    Jan 6 at 19:56




                    $begingroup$
                    Actually when I say "I discovered them", I mean that I'm the one who derived them. So as far as I'm aware, there are no papers about them. That being said, Mathematica can handle all of them.
                    $endgroup$
                    – Harmonic Sun
                    Jan 6 at 19:56












                    $begingroup$
                    @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
                    $endgroup$
                    – Jack D'Aurizio
                    Jan 7 at 0:24




                    $begingroup$
                    @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested.
                    $endgroup$
                    – Jack D'Aurizio
                    Jan 7 at 0:24


















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