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This is a curiosity question.
Let's consider the following sum: $$S=\sum_{n=0}^{+\infty} \frac{1}{\sqrt{ n !}}$$
The question asked to prove its convergence, which I did using the ratio test. So I tried to find a closed form for the this sum but without luck.
My question
1) Is there a closed form for $S$? (very likely no). 2) If the answer is no can we prove that there is no closed form for $S$ rigorously?
Thanks,
$S(a)=\displaystyle\sum_{n=0}^\infty\frac1{(n!)^a}$ converges for all $a>0$, however, the only known closed forms are $S(1)=e$,
and $S(2)=I_0(2)$, see Bessel function for more information.
IIRC there's a contour-integral expression for convolution of power series that gives the function $h(x)=f(x)\circ g(x)=\sum_{n=0}^\infty f_ng_nx^n$ in terms of the functions $f(x)=\sum_{n=0}^\infty f_nx^n$ and $g(x)=\sum_{n=0}^\infty g_nx^n$; you can invert that to get an integral equation for the power series $f(x)=\sum_{n=0}^\infty\frac{x^n}{\sqrt{n!}}$ since the convolution $f\circ f$ is equal to $e^x$. I believe it can be proven with some differential Galois theory that $f()$ is non-elementary, though, so this may not be any help in getting special values of the function. For more references, you might try Melzak's Companion to Concrete Mathematics, which uses the convolution formula to get an expression for the aforementioned Bessel functions.
