I'm looking for ideas on how to represent $\pi e$ as an infinite series of rational numbers.
It is easy to build meaningless formulas using double summations by distributing a series for $\pi$ over a series for $e$. For example, we have $$\frac{\pi e}4 = \left(\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\right)\left(\sum_{n=0}^\infty\frac1{n!}\right) = \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{(-1)^i}{(2i+1)j!}$$
This is useless, of course. An adequate answer should take form of a single sum with terms having a proper closed form.
There is a relatively simple way to build an infinite product for $\pi e$ using infinite product formulas for $\pi$ and $e$, as $$\frac{\pi e}4 = 2^{\frac12+1}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{\left(\frac12\right)^2+1}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{\left(\frac12\right)^3+1}\dots$$ but there is no easy way to turn this formula into an infinite sum.
This post discusses more infinite product formulas for $e^x$.
Another way to combine formulas for $\pi$ and $e$ is through limits. In particular, we have $$\pi = \lim_{n\to\infty}\left(\frac{2(-1)^{n+1}(2n)!}{2^{2n}B_{2n}}\right)^{1/2n}\qquad e = \lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{1/n}$$ therefore $$\pi e = \lim_{n\to\infty}\left(\frac{2(-1)^{n+1}n^{2n}}{B_{2n}}\right)^{1/2n} = \lim_{n\to\infty} n\left(\frac{2(-1)^{n+1}}{B_{2n}}\right)^{1/2n}$$ where $B_n$ stands for the $n$th Bernoulli number. This limit is not something new. It is more often written as the approximation $$B_{2n}\approx (-1)^{n+1} 4\sqrt{\pi n}\left(\frac n{\pi e}\right)^{2n}$$
It must be noted any sequence $(a_n)_n$ with limit $\pi e$ can be used in order to create the telescoping series $$\pi e = a_0+\sum_{n=0}^\infty (a_{n+1}-a_n)$$ This is a perfectly valid way of obtaining an adequate answer as long as its terms can be simplified into a simple closed form. This is not the case for the limit above, beyond its terms not being rational.
In me search I've found the following formula $$\sum_{n=0}^\infty\frac1{(2n+1)!!} = \sqrt{\frac{\pi e}2}~\mathrm{erf}\left(\frac1{\sqrt2}\right)$$
At first glance this seem useless as the error function have a $\dfrac1{\sqrt\pi}$ normalization factor, so the $\pi$ vanishes from the formula. However, the following expression due to Ramanujan fixes this issue $$\sum_{n=0}^\infty\frac1{(2n+1)!!}+\dfrac1{1+\dfrac1{1+\dfrac2{1+\dfrac3{1+\ddots}}}} = \sqrt{\frac{\pi e}2}$$ I must say this is the most astonishing formula I've ever seen, but there is still no simple way to turn this into an infinite sum, and even if there was, there would be no simple way to square it.
This is related to the continued fraction formula $$\frac{\sqrt\pi}2 e^{z^2}\mathrm{erf}(z) = \dfrac z{1-\dfrac{z^2}{\dfrac32+\dfrac{z^2}{\dfrac52-\dfrac{z^2}{\dfrac72+\ddots}}}}$$
Also, this instigated me to investigate the series $$\sum_{n=0}^\infty\frac{x^n}{\displaystyle\prod_{k=0}^n ak+b} = \frac{e^{x/a}}a\left(\frac xa\right)^{-b/a}\left(\Gamma\left(\frac ba\right)-\Gamma\left(\frac ba,\frac xa\right)\right)$$ but I couldn't get anything useful from it. Notice the double factorial series of odd numbers is the particular case of this for $x=1$, $a = 2$ and $b = 1$.
A comment pointed me out to the beautiful integral $$\int_0^1\sin(\pi x) x^x(1-x)^{1-x}dx = \frac{\pi e}{24}$$
This result and generalizations are thoroughly explored in this post. I'm doubtful about expanding this into an infinite sum, though, as there is a $\pi$ term appearing on it and the function $x^x$ doesn't have a nice series expansion.
Another difficult-to-work-with integrals I've found are the following $$\int_0^\infty\frac1{\sqrt x^{\ln x}}dx = \int_0^\infty\frac{\ln x}{\sqrt x^{\ln x}}dx = \sqrt{2\pi e}$$ $$\int_0^\infty \frac{x^{\sqrt2-1}}{x^{\ln x}}dx = \sqrt{\pi e}$$
Here is another integral mentioned in the comments
$$\int_0^\infty\frac{e^{\cos x}\sin\sin x}x dx = \frac{\pi(e-1)}2$$
This one seem to have a more promising expansion. In fact, let's notice $$\begin{aligned} e^{e^{ix}} = e^{\cos x + i\sin x} = e^{\cos x} (\cos\sin x + i\sin\sin x) \end{aligned}$$
The expansion of $e^{e^x}$ is given by $\displaystyle e\sum_{n=0}^\infty\frac{B_n}{n!}x^n$ where $B_n$ stands for the $n$th Bell number (this is not the same as Bernoulli number), therefore, we have $$e^{\cos x}\sin\sin x = e\sum_{n=0}^\infty (-1)^n\frac{B_{2n+1}}{(2n+1)!}x^{2n+1}$$ and then $$\int\frac{e^{\cos x}\sin\sin x}x = e\sum_{n=0}^\infty(-1)^n\frac{B_{2n+1}}{(2n+1)(2n+1)!}x^{2n+1}$$
Finally, because this function is $0$ at $0$, this means $$\begin{aligned} \lim_{x\to \infty}e\sum_{n=0}^\infty(-1)^n\frac{B_{2n+1}}{(2n+1)(2n+1)!}x^{2n+1} &= \frac{\pi(e-1)}2&\implies\\ \lim_{x\to \infty}\sum_{n=0}^\infty(-1)^n\frac{B_{2n+1}}{(2n+1)(2n+1)!}x^{2n+1} &= \frac\pi2-\frac\pi{2e}&\\ \end{aligned}$$ I must say, the $e$ outside the sum is very disappointing. In the end, even if we can work out the limit, we will get an expression for $\pi/e$ instead of $\pi e$. Not that that isn't impressive, though.
On an effort to contour this issue I applied Dobiński's formula $$B_n = \frac1e\sum_{k=0}^\infty\frac{k^n}{k!}$$
With it we get $$\begin{aligned} e\sum_{n=0}^\infty(-1)^n\frac{B_{2n+1}}{(2n+1)(2n+1)!}x^{2n+1} &= \sum_{n=0}^\infty\sum_{k=0}^\infty (-1)^n\frac{k^{2n+1}}{k!}\cdot\frac{x^{2n+1}}{(2n+1)(2n+1)!}\\ &= \sum_{k=0}^\infty\frac1{k!}\sum_{n=0}^\infty (-1)^n\cdot\frac{(kx)^{2n+1}}{(2n+1)(2n+1)!}\\ &= \sum_{k=0}^\infty\frac{\mathrm{Si}(kx)}{k!} \end{aligned}$$
But in this way we just end up accidentally solving the integral. In fact, all there is left to do is to notice $\displaystyle\lim_{x\to\infty}\mathrm{Si}(x) = \frac\pi2$.
A comment mentions the power series of the function $f(x) = e^{2x}\arcsin x$. Here it is, as given by Wolfram Alpha
$$x + 2x^2 + \frac{13}{6}x^3 + \frac{5}{3}x^4 + \frac{43}{40}x^5 + \frac{23}{36}x^6 + \frac{221}{560}x^7 + \frac{653}{2520}x^8 + O(x^9) = $$ $$x + \frac{4}{2!}x^2 + \frac{13}{3!}x^3 + \frac{40}{4!}x^4 + \frac{129}{5!}x^5 + \frac{460}{6!}x^6 + \frac{1989}{7!}x^7 + \frac{10448}{8!}x^8 + O(x^9) $$ OEIS have nothing to say on the numerators of these coefficients. However, there is a page on the expansion of the exponential generating function of $e^x\arctan x$, namely A279927. It gives us
$$\begin{aligned} e^x\arctan x &= \sum_{n=0}^\infty\frac{x^n}{n!}\sum_{k=0}^{n/2} (-1)^k \binom{n}{2k+1} (2k)!\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(2k+1)(n-(2k+1))!}\right)x^n\\ \end{aligned}$$
For $x = 1$ this should result in $\dfrac{\pi e}4$. Of course, this is still not the answer I'm looking for.
On the same note, A291482 is the expansion of the exponential generating function of $e^x\arcsin x$, giving
$$e^x\arcsin x = \sum_{n=0}^\infty\frac{x^n}{n!}\sum_{k=0}^{(n-1)/2} \binom{n}{2k+1} \binom{2k}{k} \frac{(2k)!}{4^k}$$
Again, for $x = 1$ this should result in $\dfrac{\pi e}2$.
The OEIS page on the decimal expansion of $\pi e$ (A019609) mentions the infinite product formula $$\pi e = \prod_{k=0}^\infty \frac{(k + 1)^{4k + 3}}{(k + 2)^{6k + 5}}\left(\frac{(2k + 3)(k + 3)}{2k + 1}\right)^{2k + 2}$$
and the limit formula $$\pi e = \lim_{k\to\infty}\frac{4k}{u_k^2}$$ where $u_k$ is defined as $u_1 = 0$, $u_2 = 1$ and $u_{k+2} = u_{k+1}+\dfrac{u_k}{2k}$. The sequence $(u_k)_k$ have generating series
$$f(x) = \sum_{k=1}^\infty u_k x^k = \frac{x^2}{(1-x)^{3/2}}e^{-x/2}$$
and closed form
$$u_k=\sum_{n=0}^{k-2}\frac{(-1)^k}{2^n n!}\binom{-3/2}{k-n-2}$$
as established in this post.
If you have any interesting formula for $\pi e$, please share. Even if it is not an infinite sum, maybe there are ways to work around it or take inspiration from it.
