# Prove that $\lim_{m\to\infty} \sum_{n=0}^{m} \frac{(-1)^n}{n!} \binom{m}{n}=0$.

binomial-coefficientscontest-mathintegrationlimitssequences-and-series

The problem from the statement is the hardest part of the problem A. 810 from KöMaL contest November 2021 (the deadline was 10 December).
After the deadline, I noticed that if $$r_0=1$$:
\begin{align} \sum_{n=0}^m r_n&=\sum_{t=0}^m\sum_{n=t}^m\frac{(-1)^t}{(t+1)!}\binom{n}t\\&= \sum_{t=0}^m\frac{(-1)^t}{(t+1)!}\sum_{n=t}^m \binom{n}{t}\\&=\sum_{t=0}^m\frac{(-1)^t}{(t+1)!}\binom{m+1}{t+1}. \end{align}
Therefore the KöMaL problem is equivalent to
\begin{align} &\lim_{m\to\infty}\sum_{n=0}^m r_n=1\Leftrightarrow\\&\lim_{m\to\infty}\sum_{t=0}^m\frac{(-1)^t}{(t+1)!}\binom{m+1}{t+1}=1\Leftrightarrow\\ &\lim_{n\to\infty}\sum_{t=1}^{m+1}\frac{(-1)^t}{t!}\binom{m+1}t=-1 \end{align}
which equivalent to

$$\displaystyle \lim_{m\to\infty} \sum_{n=0}^{m} \frac{(-1)^n}{n!} \binom{m}{n} = 0$$

and I cannot prove this affirmation.

My ideas:

1. To study the function $$f_m:\mathbb{R}\to\mathbb{R},$$ $$f_m(x)=\sum\limits_{n=0}^m (-1)^n\binom{m}{n}\frac{x^{n}}{n!}$$.
2. To use Cauchy product of two series.

Let $$I_m=\sum\limits_{n=0}^m \dfrac{(-1)^n}{n!}\dbinom{m}{n}$$ and notice that $$\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-nit} e^{e^{it}} dt = \frac{1}{n!} \tag{1}$$

This is easily proven by using the power series of the exponential and writing $$e^{e^{it}}=\sum_{k\geq 0} \frac{e^{ikt}}{k!}\tag{2}$$ and exchanging sum and integral. Then you notice that all terms but the one for $$k=n$$ are zero.

With $$(1)$$, you can then write

$$\begin{split} I_m&=\frac 1 {2\pi}\sum\limits_{n=0}^m (-1)^n\dbinom{m}{n}\int_{-\pi}^{\pi} e^{-nit} e^{e^{it}} dt\\ &= \frac 1 {2\pi}\int_{-\pi}^{\pi} \sum\limits_{n=0}^m (-1)^n\dbinom{m}{n}e^{-nit} e^{e^{it}} dt\\ &= \frac 1 {2\pi}\int_{-\pi}^{\pi} \left(1-e^{-it}\right)^m e^{e^{it}} dt \end{split}$$ With this, inspired by @Gary to sum all the $$I_m$$'s $$\sum_{m=0}^{+\infty} I_m=\frac 1 {2\pi}\int_{-\pi}^\pi \left(\sum_{m=0}^{+\infty} \left(1-e^{-it}\right)^m\right) e^{e^{it}} dt = \frac 1 {2\pi}\int_{-\pi}^\pi e^{it} e^{e^{it}} dt=0$$ where again, we get $$0$$ because you can use $$(2)$$ and observe that all terms of the series are non-constant (integer) powers of the complex exponential, which all integrate to $$0$$ over $$(-\pi, \pi)$$.

Thus $$\boxed{\lim_{m\rightarrow+\infty}I_m= 0}$$

Additional notes: My previous answer also showed that the following recurrence formula holds:

$$(m+1)I_{m+1}-2mI_m+mI_{m-1}=0$$ which can be useful for fast numerical computation of the $$I_m$$'s (each is computed in $$\mathcal O(1)$$ steps instead of $$\mathcal O(m)$$ with the original formula).

And if you consider the generating function $$f(z)=\sum_{m\geq 0}I_m z^m$$, the recurrence relation above proves that $$(z-1)^2f^\prime(z)+zf(z)=0$$, which implies that $$f(z)=\frac{C}{1-z}e^{\frac 1 {z-1}}$$ with $$C=f(0)e=e$$.