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

Solution 1:

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.

Solution 2:

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