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

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?

• Is this homework? Is there any context where the problem came up? – abiessu Jul 30 '18 at 3:48
• We know that $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!^k}$ converges for any $k>0$. So I try to evaluate the limit as $k$ approaches $0$ numerically. It seems that the limit approaches $\displaystyle \frac{1}{2}$, but I couldn't figure out why. – Star Chou Jul 30 '18 at 3:54
• You should add the background to the body of your question, so that people can see the context as they browse. Many people won't look at the comments and may very possibly vote to close your question. (Not me -- I've upvoted it.) – saulspatz Jul 30 '18 at 3:57
• Euler's transformation of alternating series may help. In this case, the first term is $1/2$ for any $k$ and perhaps you can show that the sum of remaining terms goes to $0$. – saulspatz Jul 30 '18 at 4:07

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 $\alpha = \frac{1}{2}$.

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. ////

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. ////

• My calculation suggests $\lim_{n\to\infty}\frac{\lambda_{2n+1}-\lambda_{2n}}{\lambda_{2n+2}-\lambda_{2n+1}}=1$. Should the final subscript in the denominator be $2n$ instead of $2n+1$? – J.G. Jul 30 '18 at 11:40
• @J.G. You are right, that is one crucial typo :s I fixed it now. – Sangchul Lee Jul 30 '18 at 11:45
• By the same method, for any $k\geq 1$, $$\lim_{x\rightarrow 1-} \sum_{n=0}^{\infty}(-1)^n x^{n^k}=\frac 12.$$ – i707107 Jul 30 '18 at 14:51
• @Saudman97, This is the indicator notation. For a subset $A$ of $\mathbb{R}$, we define $$\mathbf{1}_A(x) = \begin{cases} 1, & \text{if } x \in A \\ 0, & \text{if } x \notin A \end{cases}$$ – Sangchul Lee Aug 1 '18 at 5:56
• @Thinking, All this solution started with my realization that $$\sum_{n=0}^{\infty} (-1)^n e^{-\lambda_n s} = \sum_{n=0}^{\infty} \int_{\lambda_{2n}}^{\lambda_{2n+1}} s e^{-sx} \, dx$$ holds. In probabilistic interpretation, if $X$ is an exponential distribution of the rate $s$, then $$\sum_{n=0}^{\infty} (-1)^n e^{-\lambda_n s} = \sum_{n=0}^{\infty} (-1)^n \mathbb{P}[X > \lambda_n] = \mathbb{P}\left[X \in \bigcup_{n=0}^{\infty}[\lambda_{2n},\lambda_{2n+1}]\right]$$ Since $X$ will approximate "uniform distribution" over $(0,\infty)$ as $s\to0$, the limit is expected to be the density (1). – Sangchul Lee Oct 25 '18 at 0:13

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.

• Where does this proof use the fact that it's $n!$ in the denominator there rather than some other increasing function of $n$ (increasing to ensure convergence)? Are you saying it's the same if we replace $n!$ by $17^n$? Are you saying it comes out $1/2$ regardless of what's there? – Alon Amit Jul 30 '18 at 10:45
• @AlonAmit Pretty much. What we have is something whose "obvious limit" where you just substitute $x=0$ is in turn a series that "should" converge to $1/2$ in the sense that, although it doesn't converge with the usual definition of summing an infinite series, an alternative that lets it converge will give that value. I used Abel summation in this example (which makes Grandi's series $\eta(0)=1/2$), but Cesàro summation of Grandi's series does the same thing. So genuinely convergent variants on the series have to respect that. – J.G. Jul 30 '18 at 10:53
• I understand, but proving such things can be pretty subtle (e.g. Tauberian theorems). I'm not sure I see why the limit exchange there is justified, but I'm happy to think about it more. – Alon Amit Jul 30 '18 at 10:56
• @AlonAmit In light of Sangchul Lee's answer, the reason $1/2$ comes up so often is because $\lambda_{2n+2}-\lambda_{2n}\approx 2(\lambda_{2n+1}-\lambda_{2n})$ is to be expected when we linearise. – J.G. Jul 30 '18 at 11:48
• @J.G. For many other cases, $\lambda_{2n+2}-\lambda_{2n}$ may not asymptotic to $\alpha(\lambda_{2n+1}-\lambda_{2n})$. For example, $\lambda_n=\exp(n)$. – i707107 Jul 30 '18 at 14:49