Asymptotics of $\sum\limits_{n=2}^\infty \frac{x^n}{(\log n) n!}$ as $x\to\infty$ I believe, based on numerical evidence, that $$\sum_{n=2}^\infty \frac{x^n}{(\log n) n!} \sim \frac{\exp(x)}{\log(x)}$$ as $x\to\infty$. However, I am not sure how to prove this. What would be a good way to approach this problem?
 A: Consider the related finite sum
$$
s(x)=\sum\limits_n\frac{x^n}{n!}\mathbf 1_{x/2\leqslant n\leqslant 2x}=\mathrm e^x\cdot\mathbb P(x/2\leqslant N_x\leqslant 2x),
$$
where $N_x$ is a Poisson random variable with parameter $x$. A large deviation estimates ensures that, for every $x$ large enough, $\mathbb P(x/2\leqslant N_x\leqslant 2x)=1-p(x)$ where $0\lt p(x)\leqslant\mathrm e^{-cx}$ for some positive $c$. 
The series $f(x)$ to be estimated is $f(x)=g(x)+h(x)$ where $g(x)$ sums the terms such that $x/2\leqslant n\leqslant2x$ and $h(x)$ sums the other terms. Note that $0\leqslant h(x)\leqslant\mathrm e^x p(x)/\log2$ and $s(x)/\log(2x)\leqslant g(x)\leqslant s(x)/\log(x/2)$, hence
$$
\mathrm e^{x}\frac{1-\mathrm e^{-cx}}{\log x+\log2}\leqslant f(x)\leqslant\frac{\mathrm e^x}{\log x-\log2}+\mathrm e^x\frac{\mathrm e^{-cx}}{\log2}.
$$
Finally,
$$
f(x)=\frac{\mathrm e^x}{\log x}\,\left(1+O\left(\frac1{\log x}\right)\right).
$$
Note: With some more care, one can replace the $O\left(\frac1{\log x}\right)$ error term by $O\left(x^{a-1/2}\right)$, for every positive $a$.
A: Here is a possible way but is not a complete solution. (This is a bit too long for a comment and hence have community wikied it)
From Stirling, we have that
$$n! \sim \sqrt{2 \pi n} \left(\dfrac{n}{e} \right)^n$$
Equivalently, $\exp(x) \sim \sqrt{2 \pi} \dfrac{x^{x+1/2}}{x!}$. Hence, $$\exp(x) \sim \sum_{n= x- \sqrt{x}}^{x+\sqrt{x}} \dfrac{x^n}{n!}$$
(The above step needs proof.)
Hence, $$\sum_{n= x- \sqrt{x}}^{x+\sqrt{x}} \dfrac{x^n}{n! \log (n)} \sim \sum_{n= x- \sqrt{x}}^{x+\sqrt{x}} \dfrac{x^n}{n! \log (x)} \sim \dfrac{\exp(x)}{\log(x)}$$
where we have used the fact that $\log(x \pm \sqrt{x}) \sim \log(x)$.
