# Proof of this formula for $\sqrt{e\pi/2}$ and similar formulas.

\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?

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.

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

• Where can I read more about these series and other similar ones? Is there any paper or a link? Jan 6, 2019 at 19:45
• 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. Jan 6, 2019 at 19:56
• @Pinteco: the first five identities are consequences of the Poisson summation formula, if you are interested. Jan 7, 2019 at 0:24