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?
 A: 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.
A: 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.
