The function $F(x) = \sum\_{0}^{\infty} x^n/n^n$ may be familiar to many readers as an example sometimes used when teaching tests for absolute convergence of entire functions defined by power series. I know of no name for it, nor any use for it aside from pedagogical, so this is a pure curiosity question which I hope is acceptable.


The function seems to have one real zero around $x = -1.40376$; a single extremum, a minimum around $x = -5.71837$; and then to approach the $x$-axis from below asymptotically as $x$ goes to negative infinity. Is this true?



The function $\sum_{0}^{\infty} x^n/n^n$
[**Edited** *to outline the end of the argument that $f(-M) \rightarrow 0$
(and to correct a few typos etc. while I'm at it)*]


Yes, $F(x) \rightarrow 0$ from below as $x \rightarrow -\infty$.
The convergence is slow, and precise asymptotic analysis seems to be
 somewhat annoying because it involves the lower branch of the
[Lambert W function](http://en.wikipedia.org/wiki/Lambert_W_function).


The massive cancellations in $\sum\_{n=0}^\infty x^n/n^n$ for
$x \rightarrow -\infty$ can be tamed by the familiar device of writing
$$
\frac1{n^n} = \frac1{(n-1)!} \int\_0^\infty t^{n-1} e^{-nt} dt
$$
for $n=1,2,3,\ldots$. Multiplying by $x^n$, summing over $n>0$,
and restoring the $n=0$ term 
$x^0/0^0=1$ yields
$$
f(x) = 1 + x \int\_0^\infty e^{txe^{-t}} e^{-t} dt.
$$
Hence if $x=-M$ then
$$
f(x) = f(-M) = 1 - M \int\_0^\infty e^{-Mte^{-t}} e^{-t} dt,
$$
and as $M \rightarrow +\infty$
the integral naturally splits into the parts $t \leq 1$ where
$t e^{-t}$ is increasing and $t \geq 1$ where $t e^{-t}$ is decreasing.
We let $u = t e^{-t}$, so the integrand becomes
$e^{-Mu} du/(1-t)$. For $t<1$ we use
[Abel's power series](http://math.harvard.edu/~elkies/Misc/abel.pdf)
$t = \sum\_{m=1}^\infty m^{m-1} u^m/m!$ to expand the integral in
an asymptotic series:
$$
\int\_0^1 e^{-Mte^{-t}} e^{-t} dt \sim
 \frac1M + \frac1{M^2} + \frac{2^2}{M^3} + \frac{3^3}{M^4} + \frac{4^4}{M^5} + \cdots
$$
which is already enough to get $f(-M) < 0$ for large $M$.
[Curiously the asymptotics of $\sum\_{n=0}^\infty (-M)^n/n^n$
have led us to the divergent series $\sum\_{n=1}^\infty n^n/M^n$.]


But the resulting bound $f(-M) < -1/M$ underestimates $|f(-M)|$: numerically
$f(-100) \simeq -.1826$, $\phantom.$ $f(-1000) \simeq -.1180$, and
$\phantom.$ $f(-10000) \simeq -.0899$, suggesting that $f(-M)$ decays
only as $-1/\log M$ or so. The reason must be the $t>1$ part of the integral.
On this part, $t = \log(1/u) + \log\log(1/u) + o(1)$ as $t \rightarrow \infty$,
so the integral behaves to first order like
$\int\_0^{1/e} e^{-Mu} du / \log(1/u)$. Now $\log(1/u) \rightarrow 0$
as $u \rightarrow 0+$, but the convergence is slower than any positive
power of $u$. Therefore, the integral is $o(1/M)$, which completes
the argument that $f(x) \rightarrow 0$ as $x \rightarrow -\infty$;
but the integral is not $O(1/M^\theta)$ for any $\theta > 1$, so $f(-M)$
decays slower than any positive power of $M$.


A more thorough asymptotic analysis of the $t>1$ integral as
$M \rightarrow \infty$ looks routine but unpleasant, so I'll stop
at this point; perhaps somebody else here will be interested in
pursuing it further.


