The value of a limit of a power series: $\lim\limits_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$ What is the answer to the following limit of a power series?
$$\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$$
 A: A simple calculation shows that
\begin{align*}
\sum_{k=1}^{\infty} (-1)^{k} \left( \frac{x}{k} \right)^{k}
&= \sum_{k=1}^{\infty} \frac{(-1)^{k} x^{k}}{(k-1)!} \int_{0}^{\infty} t^{k-1} e^{-kt} \, dt 
= \int_{0}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k} x^{k} t^{k-1} e^{-kt}}{(k-1)!} \, dt \\
&= -x \int_{0}^{\infty} \exp \left\{ - t \left( 1 + x e^{-t} \right) \right\} \, dt
= - \int_{0}^{1} x \cdot u^{x u} \, du,
\end{align*}
where $u = e^{-t}$. Now we claim that
$$ \lim_{x\to\infty} \int_{0}^{1} x \cdot u^{x u} \, du = 1. $$
To find the limit, we prove the following lemma:

Lemma. Let $f : [0, \delta] \to [0, 1]$ be a measurable function. Suppose there exists $0 < A < B$ such that
  $$ 1 - Ax \leq f(x) \leq 1 - Bx. $$
  Then we have
  $$ \frac{1}{A} \leq \liminf_{x\to\infty} \left( x \int_{0}^{\delta} f(t)^{x} \, dt \right) \leq \limsup_{x\to\infty} \left( x \int_{0}^{\delta} f(t)^{x} \, dt \right) \leq \frac{1}{B}. $$

Assume this lemma holds. Let $f(u) = u^{u}$. Then we observe that


*

*$f(u)$ decreases for $[0, 1/e]$ and increases for $[1/e, 1]$.

*For any small $\epsilon > 0$, there exists small $\delta > 0$ such that
$$ 1 - (1+\epsilon)(1-u) \leq f(u) \leq 1 - (1-\epsilon)(1-u) $$
for $0 < u < \delta$.

*For any large $M > 0$, we can choose small $\delta > 0$ such that
$$f'(u) = u^{u}(1 + \log u) \leq -M$$
for $0 < x < \delta$. In particular, $f(u) \leq 1 - Mu$.


Let $\epsilon > 0$ be small and $M > 0$ be large. Let $\delta > 0$ be a sufficiently small number satisfying the conditions simultaneously. Then we have
$$ 0 \leq x \int_{\delta}^{1-\delta} u^{xu} \, du \leq x \max\{ f(\delta)^{x}, f(1-\delta)^{x} \} \xrightarrow{x\to\infty} 0. $$
Also, Lemma shows that
$$ \frac{1}{1+\epsilon} \leq \liminf_{x\to\infty} \left( x \int_{1-\delta}^{1} u^{xu} \, du \right) \leq \limsup_{x\to\infty} \left( x \int_{1-\delta}^{1} u^{xu} \, du \right) \leq \frac{1}{1-\epsilon} $$
and
$$ 0 \leq \limsup_{x\to\infty} \left( x \int_{0}^{\delta} u^{xu} \, du \right) \leq \frac{1}{M}. $$
Putting together, we have
$$ \frac{1}{1+\epsilon} \leq \liminf_{x\to\infty} \left( x \int_{0}^{1} u^{xu} \, du \right) \leq \limsup_{x\to\infty} \left( x \int_{0}^{1} u^{xu} \, du \right) \leq \frac{1}{1-\epsilon} + \frac{1}{M}. $$
Therefore, letting $M \to \infty$ and $\epsilon \to 0^{+}$, we obtain
$$ \lim_{x\to\infty} \left( x \int_{0}^{1} u^{xu} \, du \right) = 1 $$
as desired.

Proof of Lemma. For any $0 < \eta < \delta$, we have
  $$ 0 \leq x \int_{\eta}^{\delta} f(t)^{x} \, dt \leq x \int_{\eta}^{\delta} \max \{ 1- B\eta, 0 \}^{x} \, dt \leq \max \{ x \delta (1- B\eta)^{x}, 0 \} \xrightarrow[]{x\to\infty} 0. $$
  Thus we may assume that $\delta$ is sufficiently small so that $1 - A\delta \geq 0$. Then
  \begin{align*}
x \int_{0}^{1/A} (1 - At)^{x} \, dt + o(1)
&= x \int_{0}^{\delta} (1 - At)^{x} \, dt \\
&\leq x \int_{0}^{\delta} f(t)^{x} \, dt \\
&\leq x \int_{0}^{\delta} (1 - Bt)^{x} \, dt = \leq x \int_{0}^{1/B} (1 - Bt)^{x} \, dt + o(1).
\end{align*}
  Evaluating, we obtain
  $$ \frac{x}{A(x+1)} + o(1) \leq x \int_{0}^{\delta} f(t)^{x} \, dt \leq \frac{x}{B(x+1)} + o(1), $$
  proving the lemma.

A: 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, 
$$\begin{eqnarray*}
\lim_{x\to\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k 
&=& -\lim_{x\to\infty} x \int_0^\infty e^{-z(xe^{-z}+1)} dz \\
&=& -\lim_{x\to\infty} (I_1+I_2) \\
&=& -1.
\end{eqnarray*}$$
