Sum : $f(x) = \sum_n \frac{x^n}{\sqrt{n!}}$. Is it possible to find a sum of a series:
$$f(x) = \sum_n \frac{x^n}{\sqrt{n!}}$$
at least as a combination of some special functions? 
 A: As already pointed out in the comments the function $f(x)$ does not have a closed form, however I'll post here my way of finding its asymptotic behaviour at $x \to \infty$, in case it is useful to anyone. 
$$f'(x) = \frac{1}{x} \sum_n \frac{x^n }{\sqrt{n!}}n.$$
Suppose that $f'(x) =\frac{1}{x}g(x) f(x)$.
$$g(x)=\frac{\sum \frac{x^n }{\sqrt{n!}}n}{\sum \frac{x^n }{\sqrt{n!}}}$$
Using the fact that:
$$ \lim_{x\to \infty} \frac{a_n x^n}{b_n x^n} =\lim_{n\to \infty} \frac{a_n}{b_n},$$
$$\lim_{x\to \infty} \frac{g(x)}{x^2}=\frac{\sum \frac{x^n }{\sqrt{n!}}n}{\sum \frac{x^{n+2} }{\sqrt{n!}}} = \lim_{n\to \infty} \frac{n}{\sqrt{n(n-1)}} =1$$
Therefore $g(x) = x^2 + o(x^2)$.
$$\lim_{x\to \infty} g(x) - x^2 =\frac{\sum \frac{x^n }{\sqrt{n!}}n -\sum \frac{x^{n+2} }{\sqrt{n!}} }{\sum \frac{x^{n} }{\sqrt{n!}}} = \lim_{n\to \infty} n -\sqrt{n(n-1)} = \frac{1}{2}$$
Therefore $g(x) = x^2 + \frac{1}{2}+o(1)$. Now we have a simple differential equation:
$$f' = \left(x + \frac{1}{2x}\right)f,$$ which gives the answer:
$$f \sim \sqrt{x}e^{\frac{x^2}{2}},  \, x\to \infty$$ 
Next terms of the approximation of $g(x)$ will be $\sim c/x^k$, $k>1$ which gives multipliers of $\exp(c/x^{k+1})$ in $f$, which quickly goes to 1.
