Behaviour of $\sum_{n=1} \frac{1}{n!} e^{an}\log(n)$ as $a \to \infty$ 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.
 A: We expect the largest summand approximately where $n\approx e^a$.
Comparing the log to its tangent at that point, 
$$\ln n\le  a-1+\frac{n}{e^a}$$
So
$$\begin{align}\sum_{n=1}^\infty\frac1{n!}e^{an}\ln n&\le (a-1)\sum_{n=1}^\infty\frac1{n!}e^{an}+ \sum_{n=1}^\infty\frac1{(n-1)!}e^{a(n-1)}\\&=(a-1)(e^{e^a}-1)+e^{e^a}\\&=e^{\ln a+e^a}-a+1\end{align}$$
A: 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.
A: Let $N(t)$ be the Poisson process of the unit rate. Then
$$ \sum_{n=0}^{\infty} \frac{e^{an}}{n!}f(n) = e^{e^a}\mathbb{E}[f(N(e^a))]. $$
In our case, one can set $f(n) = \log(n \vee 1)$. In view of the law of large numbers, one may expect that this is approximately
$$e^{e^a}f(\mathbb{E}[N(e^a)]) = e^{e^a}f(e^a) = a e^{e^a}.$$

Indeed, by the Bennett-type concentration inequality, with $\psi(x) = (1+x)\log(1+x)-x $, for any $\epsilon > 0$ we have
$$ \mathbb{P}( \left| N(\lambda) - \lambda \right| \geq \epsilon\lambda) \leq 2e^{-\psi(\epsilon)\lambda}. $$
Then by using the inequality $\log(\lambda \pm a) \leq \log(\lambda+|a|)=\log\lambda + \int_{0}^{|a|/\lambda} \frac{ds}{1+s} $, we have
\begin{align*}
\mathbb{E} \left[ \log(N(\lambda)\vee 1) \mathbf{1}_{\{|N(\lambda)-\lambda|\geq\epsilon\lambda\}} \right]
&\leq \mathbb{E} \left[ \log \left( \lambda + \left|N(\lambda) - \lambda\right| \right) \mathbf{1}_{\{|N(\lambda)-\lambda|\geq\epsilon\lambda\}} \right] \\
&\leq (\log\lambda)\mathbb{P}( \epsilon\lambda \leq \left|N(\lambda) - \lambda\right|) \\
&\quad + \int_{0}^{\infty} \frac{ds}{1+s} \mathbb{P}( (s\vee\epsilon)\lambda \leq \left|N(\lambda) - \lambda\right|) \\
&\leq 2(\log \lambda)e^{-\psi(\epsilon)\lambda} + \int_{0}^{\infty} \frac{2e^{-\psi(s\vee \epsilon)\lambda}}{1+s}\,ds 
\end{align*}
This bound vanishes as $\lambda \to \infty$, so with $f(n) = \log(n\vee 1)$, it follows that
$$ \lim_{\lambda\to\infty} \frac{\mathbb{E} \left[ f(N(\lambda)) \right] }{\log\lambda} = 1 $$
This proves the desired asymptotics:
$$ \lim_{a\to\infty} \frac{\sum_{n=1}^{\infty} \frac{1}{n!}e^{an}\log n}{a e^{e^a}} =\lim_{a\to\infty} \frac{\mathbb{E}[f(N(e^a))]}{a} = 1. $$
