TITLE: Behaviour of $\sum_{n=1} \frac{1}{n!} e^{an}\log(n)$ as $a \to \infty$
QUESTION [7 upvotes]: I am interested in the behavior of the following sum
\begin{align}
\sum_{n=1} \frac{1}{n!} e^{an}\log(n),
\end{align}
as $a$ approaches infinity. 
Using $\log(n) \le n$ we can the following upper bound
\begin{align}
\sum_{n=1} \frac{1}{n!} e^{an}\log(n) \le   e^{a+e^a}. 
\end{align}
However, I think the true behavior is more like  $e^{\log(a)+e^a}$. However, I am not sure how to show whether this is true or not.

REPLY [4 votes]: A full asymptotic series for $S$, with $x=\exp(a) \to \infty,$ 
$$ S:=\sum_{n=1}^\infty \frac{x^n}{n!} \log{n} \sim e^x\big(\, \log(x) - \frac{1}{2x}-\frac{5}{12x^2} + ... \big)$$
can be derived in the following manner. 
It is a well-known Laplace transform that can be rearranged as such:
$$ \frac{\log{n}}{n} =-\frac{\gamma}{n} - \int_0^\infty e^{-nt} \,\log{t}\, dt. $$
The $\gamma$ is Euler's constant. Insert this into definition for S, interchange order of sum and integration, and sum the series to get
$$ S=-\gamma\big(e^x-1)-x\int_0^\infty e^{-t}\exp{(x\, e^{-t})} \log{t} \, dt. $$
This exact representation takes a simpler form on the substitution $e^{-t} \to u,$
$$ S=-\gamma\big(e^x-1)-x\int_0^1 \exp{(x\,u)} \log{(\log{(1/u)})} \, du. $$
Clearly for $x$ large what matters the most is what happens near $u=1.$ Expand the log-log as
$$ \log{(\log{(1/u)})} = \log{(1-u)} - \frac{1}{2}(u-1) + \frac{5}{24}(u-1)^2+... $$
Now
$$\int_0^1\log{(1-u)}\exp(x\,u)du=-\frac{e^x}{x}\big(\gamma+\Gamma(0,x)+\log(x)\big) \sim  -\frac{e^x}{x}\big(\gamma+\log(x)\big)$$
where the incomplete gamma function is neglected because its asymptotic expansion is exponentially small in comparison. Likewise, neglecting terms that are exponentially small, we find that
$$ \int_0^1(u-1)\exp(x\,u) du \sim -\frac{e^x}{x^2} \text{ and } 
 \int_0^1(u-1)^2\exp(x\,u) du \sim \frac{2e^x}{x^3} .$$
Algebra completes the solution.