TITLE: Asymptotics of the sum $\sum_{n=1}^\infty \frac{x^n}{n^n}$
QUESTION [6 upvotes]: How does the sum
$$f(x)=\sum_{n=1}^\infty \frac{x^n}{n^n}$$
behave asymptotically as $x\to\infty$? It appears that $f(x)$ asymptotically dominates any polynomial and is dominated by any exponential, so we might consider $\log f(x)$ rather than $f(x)$.
I apologize for having no work to show on this problem; I have no idea how to begin tackling a problem regarding the asymptotics of a function given its power series (of which there is no hope of evaluating in closed form). Hopefully an answer will provide me with some tools for doing so.

REPLY [2 votes]: Similar to Maxim: 
$$
\begin{align}
f(a)&=\sum_{n=1}^\infty \frac{a^n}{n^n}=\sum_{n=1}^\infty e^{n \log(a/n)}\\
&\approx \int_1^\infty e^{t \log(a/t)} dt \\
&= a \int_0^a \frac{1}{u^2} e^{a \log(u) /u} du\\
&= a \int_0^a h(u) e^{a g(u)} du\\
&\approx a \sqrt{\frac{2 \pi}{a |g''(u_0)|} } h(u_0) e^{a g(u_0)}
\end{align}
$$
where we've used Laplace's approximation (assuming $a \gg e$) to $h(u) =\frac{1}{u^2}$ and $g(u)=\log(u)/u$, with $u_0=e$ , $g''(u_0)=-1/e^3$ . Then the approximation gives
$$f(a)\approx \sqrt{2 \pi a} \exp( a/e-1/2)$$
or 
$$\log f(a)\approx  \frac{a}{e} + \frac{1}{2}\log(a) + \frac{1}{2}(\log(2 \pi)-1) $$
I've not done the strict asyptotical analysis, but it looks as if the error is $o(1)$. Some numerical values
 a      log(f(a))    aprox      abs error 
3.0     1.896554    2.071883    0.175329
5.0     2.984687    3.063055    0.078368
7.5     4.150135    4.185486    0.035350
10.0    5.229637    5.249025    0.019389
20.0    9.268130    9.274393    0.006264
30.0    13.151944   13.155920   0.003976
50.0    20.766590   20.768922   0.002332
75.0    30.167102   30.168641   0.001539
100.0   39.508319   39.509468   0.001149
200.0   76.643415   76.643985   0.000570
300.0   113.634283  113.634662  0.000379
500.0   187.465736  187.465963  0.000227
750.0   279.638405  279.638556  0.000151
1000.0  371.752144  371.752257  0.000113