Can we find a closed form for this infinite sum?
$$\sum_{k=1}^\infty{_{0}F_1}(;2;-kx)$$
This arises from the problem of finding a closed form for
$$\sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{\zeta\left(2k\right)}{\left(2k\right)!}h^{2k-1}.$$
Here was my process. Begin with the special form of the Euler-Maclaurin formula.
$$ \begin{align} \sum_{k=1}^n \frac{h^{2k-1}B_{2k}}{(2k)!} \left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right) &=\sum_{k=1}^n f(kh+a)-\left(\frac{f(b)-f(a)}{2}\right) \\&-\frac{1}{h}\int_a^b f(t)dt \\&-R \end{align} $$
where $h=\frac{b-a}{n}$. However, $h$ becomes unrestricted if we let $b\to\infty$ and $n\to\infty$. We also may limit $a\to 0$ to simplify things.
Consider the Bessel Functions $J_n(x)$ and calculate via Taylor series
$$\lim\limits_{x\to 0}\frac{d^m}{dx^m}J_0\left(2\sqrt{x}\right)=\frac{(-1)^m}{m!}.$$
Then we have
$$\lim\limits_{x\to 0}\frac{d^{m-1}}{dx^{m-1}}\frac{J_1\left(2\sqrt{x}\right)}{\sqrt{x}}=\lim\limits_{x\to 0}\frac{d^{m-1}}{dx^{m-1}}{_{0}F_1(;2;-x)}=\frac{(-1)^m}{m!}$$
and
$$\lim\limits_{x\to \infty}\frac{d^m}{dx^m}{_{0}F_1(;2;-x)}=0.$$
Then since
$$\frac{(2\pi)^{2k}B_{2k}}{2(2k)!}=(-1)^{k-1}\zeta(2k),$$
the Euler-Maclaurin formula yields
$$ \begin{align} \sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{\zeta\left(2k\right)}{\left(2k\right)!}h^{2k-1} &=\frac{\pi}{2}-\frac{1}{2h}+\pi\sum_{k=1}^{\infty}\frac{J_1\left(2\sqrt{2\pi hk}\right)}{\sqrt{2\pi hk}}\\ &=\frac{\pi}{2}-\frac{1}{2h}+\pi\sum_{k=1}^{\infty}{_{0}F_1(;2;-2\pi hk)} \end{align}. $$
To make life simpler (at least mine), I prefer to use $$\, _0F_1(;2;-k x)=\frac{J_1\left(2 \sqrt{k} \sqrt{x}\right)}{\sqrt{k} \sqrt{x}}$$ $$\sum_{k=1}^\infty{_{0}F_1}(;2;-kx)=\frac 1{\sqrt{x}}\sum_{k=1}^\infty \frac 1{\sqrt{k}}J_1\left(2 \sqrt{k} \sqrt{x}\right)$$ By the integral test, the summation converges since $$\int_1^\infty \frac 1{\sqrt{k}}J_1\left(2 \sqrt{k} \sqrt{x}\right)\,dk=\frac{J_0\left(2 \sqrt{x}\right)}{\sqrt{x}}$$
For the summation, I am stuck.