TITLE: The value of a limit of a power series: $\lim\limits_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$
QUESTION [7 upvotes]: What is the answer to the following limit of a power series?
$$\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$$

REPLY [2 votes]: It is straightforward to reproduce the integral representations of the sum found by @deoxygerbe and @user17762 in their interesting paper and by @sos440,
$$\begin{eqnarray*}
\sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k
&=& -x\int_0^1 t^{x t} dt \\
&=& -x \int_0^\infty e^{-z(xe^{-z}+1)} dz.
\end{eqnarray*}$$
Here we verify @sos440's result, that the sum is $-1$, by examining the last integral.
We have 
$$\begin{eqnarray*}
x \int_0^\infty e^{-z(xe^{-z}+1)} dz 
&=& \underbrace{\int_0^1 f(z)dz}_{I_1} + \underbrace{\int_1^\infty f(z)dz}_{I_2},
\end{eqnarray*}$$
where $f(z) = x e^{-z(xe^{-z}+1)}$.
On $(1,\infty)$, $f(z)$ has a global max at $z_0$, the solution to
$x(z_0-1)=e^{z_0}$.
Note that $z_0 \sim \log x + \log\log x$
so that $\lim_{x\to\infty} z_0 = \infty$.
Applying Laplace's method we find
$$\begin{eqnarray*}
I_2
&\sim& \sqrt{\frac{2\pi}{(z_0-2)(z_0-1)}} e^{-z_0/(z_0-1)} \\
&\sim& \frac{\sqrt{2\pi}}{e z_0}.
\end{eqnarray*}$$
Thus, in the limit, $I_2 = 0$. 
But
$$\begin{array}{ccccc}
\displaystyle x\int_0^1 e^{-z(x+1)} dz
&\le&
\displaystyle x \int_0^1 e^{-z(xe^{-z}+1)} dz
&\le& 
\displaystyle x\int_0^1 e^{-z[x(1-z)+1]} dz.
\end{array}$$
(Here we use that on $[0,1]$, $1-z\le e^{-z}\le 1$.) 
The bounding integrals can be calculated explicitly and in the limit they are unity. 
Thus, in the limit, $I_1 = 1$. 
Therefore, 
$$\begin{eqnarray*}
\lim_{x\to\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k 
&=& -\lim_{x\to\infty} x \int_0^\infty e^{-z(xe^{-z}+1)} dz \\
&=& -\lim_{x\to\infty} (I_1+I_2) \\
&=& -1.
\end{eqnarray*}$$