\begin{align} \sqrt{\frac{e\pi}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\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?


2 Answers 2


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.

  • $\begingroup$ The first addend (the series) can also be expressed as a nice continued fraction. Kindly see this post. $\endgroup$ Oct 22, 2023 at 14:03
  • $\begingroup$ And apparently, there's a cubic version of the identity as discussed in this new post. $\endgroup$ Oct 23, 2023 at 11:18

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

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

$\cdot$ The Stirling limit : $$\lim_{n\to\infty}\frac{n!e^n}{n^n\sqrt{n}}=\sqrt{2\pi}$$

$\cdot$ 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.

  • $\begingroup$ Where can I read more about these series and other similar ones? Is there any paper or a link? $\endgroup$
    – Pinteco
    Jan 6, 2019 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. $\endgroup$ Jan 6, 2019 at 19:56
  • 2
    $\begingroup$ @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested. $\endgroup$ Jan 7, 2019 at 0:24

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