Proving $\lim_{x \to 0+} \sum_{n=0}^\infty \frac{(-1)^n}{n!^x} = \frac{1}{2}$

Question

Prove that $$ \lim_{x \to 0+} \sum_{n=0}^\infty \frac{(-1)^n}{n!^x} = \frac{1}{2}. $$

We know that $$ \sum_{n=0}^\infty \frac{(-1)^n}{n!^x}$$ converges for any $x>0$. So I try to evaluate the limit as $x$ approaches $0$ numerically. It seems that the limit approaches $\displaystyle \frac{1}{2}$.

I know that $$\sum_{n=0}^\infty \frac{(-1)^n}{n!} = \frac{1}{e}.$$

Does it help to solve this problem?

Answer 1

Our main claim is as follows:

Proposition. Let $(\lambda_n)$ be an increasing sequence of positive real numbers. If $(\lambda_n)$ satisfies $$\lim_{R\to\infty} \frac{1}{R} \int_{0}^{R} \sum_{n=0}^{\infty} \mathbf{1}_{[\lambda_{2n}, \lambda_{2n+1}]}(x) \, dx = \alpha \tag{1} $$ for some $\alpha \in [0, 1]$, then $$\lim_{s\to0^+} \sum_{n=0}^{\infty} (-1)^n e^{-\lambda_n s} = \alpha \tag{2} $$

Here, a sequence $(\lambda_n)$ is increasing if $\lambda_n \leq \lambda_{n+1}$ for all $n$. As a corollary of this proposition, we obtain the following easier criterion.

Corollary. Let $(\lambda_n)$ be an increasing sequence of positive real numbers that satisfy

1. $\lim_{n\to\infty} \lambda_n = \infty$,
2. $\lim_{n\to\infty} \lambda_{n+1}/\lambda_n = 1$,
3. $\lambda_{2n} < \lambda_{2n+2}$ hold for all sufficiently large $n$ and $$ \lim_{n\to\infty} \frac{\lambda_{2n+1} - \lambda_{2n}}{\lambda_{2n+2} - \lambda_{2n}} = \alpha. \tag{3} $$

Then we have $\text{(1)}$. In particular, the conclusion $\text{(2)}$ of the main claim continues to hold.

Here are some examples:

• The choice $\lambda_n = \log(n+1)$ satisfies the assumptions with $\alpha = \frac{1}{2}$. In fact, this reduces to the archetypal example $\eta(0) = \frac{1}{2}$.

• OP's conjecture is covered by the corollary by choosing $\lambda_n = \log(n!)$ and noting that $\text{(3)}$ holds with $\alpha = \frac{1}{2}$.

• If $P$ is a non-constant polynomial such that $\lambda_n = P(n)$ is positive, then $(\lambda_n)$ must be strictly increasing for large $n$, and using the mean value theorem we find that the assumptions are satisfied with $\alpha = \frac{1}{2}$.

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Proof of Proposition. Write $F(x) = \int_{0}^{x} \left( \sum_{n=0}^{\infty} \mathbf{1}_{[\lambda_{2n}, \lambda_{2n+1}]}(t) \right) \, dt$ and note that

\begin{align*} \sum_{n=0}^{\infty} (-1)^n e^{-\lambda_n s} &= \sum_{n=0}^{\infty} \int_{\lambda_{2n}}^{\lambda_{2n+1}} s e^{-sx} \, dx = \int_{0}^{\infty} s e^{-sx} \, dF(x) \\ &= \int_{0}^{\infty} s^2 e^{-sx} F(x) \, dx \stackrel{u=sx}{=} \int_{0}^{\infty} s F(u/s) e^{-u} \, du. \end{align*}

Since $0 \leq F(x) \leq x$, the integrand of the last integral is dominated by $ue^{-u}$ uniformly in $s > 0$. Also, by the assupmption $\text{(1)}$, we have $s F(u/s) \to \alpha u$ as $s \to 0^+$ for each $u > 0$. Therefore, it follows from the dominated convergence theorem that

$$ \lim_{s\to0^+} \sum_{n=0}^{\infty} (-1)^n e^{-\lambda_n s} = \int_{0}^{\infty} \alpha u e^{-u} \, du = \alpha, $$

which completes the proof. $\square$

Proof of Corollary. For each large $R$, pick $N$ such that $\lambda_{2N} \leq R \leq \lambda_{2N+2}$. Then

$$ \frac{1}{R} \int_{0}^{R} \sum_{n=0}^{\infty} \mathbf{1}_{[\lambda_{2n}, \lambda_{2n+1}]}(x) \, dx \leq \frac{\lambda_{2N+2}}{\lambda_{2N}} \cdot \frac{\sum_{n=0}^{N} (\lambda_{2n+1} - \lambda_{2n})}{\sum_{n=0}^{N} (\lambda_{2n+2} - \lambda_{2n})} $$

and this upper bound converges to $\alpha$ as $N\to\infty$ by Stolz–Cesàro theorem. Similar argument applied to the lower bound

$$ \frac{1}{R} \int_{0}^{R} \sum_{n=0}^{\infty} \mathbf{1}_{[\lambda_{2n}, \lambda_{2n+1}]}(x) \, dx \geq \frac{\lambda_{2N}}{\lambda_{2N+2}} \cdot \frac{\sum_{n=0}^{N-1} (\lambda_{2n+1} - \lambda_{2n})}{\sum_{n=0}^{N-1} (\lambda_{2n+2} - \lambda_{2n})} $$

proves the desired claim together with the squeezing theorem. $\square$

Answer 2

Define $S(x,\,y):=\sum_{n\ge 0}\frac{(-1)^n}{n!^x}e^{-ny}$, which converges for any $x>0$ with $y\ge 0$ and any $y>0$ with $x\ge 0$. Grandi's series $\sum_{n\ge 0}(-1)^n$ doesn't converge to any specific value (although its partial sums also don't tend to $\pm\infty$ either), but it is said to Abel summable to $\frac{1}{2}$ in the sense $\lim_{y\to 0^+}S(0,\,y)=\frac{1}{2}$, which you can easily prove with geometric series. The proof you're looking for is $$\lim_{x\to 0^+}S(x,\,0)=\lim_{x\to 0^+}\lim_{y\to 0^+}S(x,\,y)=\lim_{y\to 0^+}\lim_{x\to 0^+}S(x,\,y)=\lim_{y\to 0^+}S(0,\,y)=\frac{1}{2}.$$The part that requires a careful explanation is why we can commute the limits at the second $=$ sign. Again, the key insight is that the leftmost limit is computed for a non-zero argument, and that implies varying the rightmost limit towards $0$ has it continuously converge to a finite value.