Asymptotics of the sum $\sum_{n=1}^\infty \frac{x^n}{n^n}$ 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.
 A: If we apply the identity
$$
\frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} \mathrm{e}^{ - nt} \mathrm{d}t} , \quad n\geq 1,
$$
we obtain the integral representation
$$
\sum\limits_{n = 1}^\infty  {\frac{{x^n }}{{n^n }}}  = x\int_0^{ + \infty } {\mathrm{e}^{ - xp(t)} q(t) \mathrm{d}t} ,
$$
with $p(t) =  - t\mathrm{e}^{ - t}$ and $q(t)=\mathrm{e}^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find
$$
\sum\limits_{n = 1}^\infty  {\frac{{x^n }}{{n^n }}}  \sim \mathrm{e}^{z} \sqrt {2\pi z} \sum\limits_{k = 0}^\infty  {\frac{{a_k }}{{z^k }}}  = \mathrm{e}^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right)
$$
as $x\to +\infty$, with $z=x/\mathrm{e}$ and
$$
a_k  = \frac{1}{{2^k k!}}\left[ {\frac{{\mathrm{d}^{2k} }}{{\mathrm{d}t^{2k} }}\left( {\mathrm{e}^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)\mathrm{e}^{ - t} }}} \right)^{k + 1/2} } \right)} \right]_{t = 0} .
$$
An alternative expression for the coefficients involving the Bernoulli polynomials $B_j(x)$ is
$$
a_k  = \frac{1}{k!}\left[ \frac{\mathrm{d}^k }{\mathrm{d}t^k}\exp \left( \sum\limits_{j = 1}^k B_{j + 1}\! \left( \tfrac{3}{2} - k \right)\frac{t^j}{j(j + 1)} \right) \right]_{t = 0} .
$$
I omit the proof.
A: Here is a simple argument that gives simple bounds. The idea is just to apply Stirling's approximation in reverse. Namely, using the crude Stirling inequalities
$$e \left( \frac{n}{e} \right)^n \le n! \le en \left( \frac{n}{e} \right)^n$$
we get the inequalities
$$e^{n-1} (n-1)! \le n^n \le e^{n-1} n!$$
which give, for $x \ge 0$, the upper and lower bounds
$$\boxed{ e^{\frac{x}{e} + 1} - e = \sum_{n \ge 1} \frac{x^n}{e^{n-1} n!} \le f(x) \le \sum_{n \ge 1} \frac{x^n}{e^{n-1} (n-1)!} = x e^{\frac{x}{e}} }.$$
This is already enough to establish the asymptotic behavior of $f(x)$ as $x \to \infty$ up to a factor of $x$, which in particular gives $\boxed{ \ln f(x) \sim \frac{x}{e} + O(\ln x) }$.
A: There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$, computing the expansion of $n'(\xi)$ at $\xi = 0$ and extending the integration range to $(-\infty, \infty)$, we obtain
$$\frac {n'(\xi)} x =
\sqrt{\frac 2 e} + c_1 \xi -
 \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4),
 \quad \xi \to 0,\\
\sum_{n \geq 1} \frac {x^n} {n^n} =
\int_{-\infty}^\infty
 x \left( \sqrt{\frac 2 e} - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 +
  O(\xi^4) \right)
 e^{x/e - x \xi^2} d\xi = \\
\sqrt{\frac \pi 2} \,e^{x/e} \left(
 2\sqrt{\frac x e} - \frac 1 {12} \sqrt{\frac e x} + O(x^{-3/2}) \right),
 \quad x \to \infty,$$
which gives $\ln f(x)$ with an error of order $O(x^{-2})$.
