The value of a limit of a power series: $limlimits_{xrightarrow +infty} sum_{k=1}^infty (-1)^k left(frac{x}{k} right)^k$

What is the answer to the following limit of a power series?

$$\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$$

Answer

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

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


2. 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 $0 < u < \delta$.


3. 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.