Is $\sum_{n=0}^\infty \frac{(-1)^n}{n!(x-n)!}$ equal to $0$? I would like to show that  $\sum_{n=0}^\infty \frac{(-1)^n}{n!(x-n)!}=0$ for $x\in\mathbb{R}$, x>0. Proving this for odd numbers is easy, it should't be hard for even numbers too (However I managed to prove it only for odd numbers), becouse when $x$ is natural then the sum becomes finite $\sum_{n=0}^x \frac{(-1)^n}{n!(x-n)!}$. The real problem is to prove this for all real numbers (Assuming $n!=\Gamma(n+1)$.). I was thinking about using Euler's reflection formula to avoid factorials in the sum, but I didn't come to anything nice.
Thanks for all the help.
 A: If $x=0$ the result is false since the sum is equal to $1$ (the term with $n=0$ remains). So assume that $x>0$. We have, by the Newton binomial theorem, that$$\frac{1}{x!}\sum_{n\geq0}\dbinom{x}{n}\left(-1\right)^{n}=\frac{\left(1-1\right)^{x}}{x!}=0$$ and so the thesis.
A: Let $ m $ be a positive integer, and $ x\in\left]0,+\infty\right[ $, we have the following : $$ \left(\forall n\in\mathbb{N}\right),\ \binom{x}{n}=\binom{x-1}{n}+\binom{x-1}{n-1} $$
Thus : \begin{aligned} \sum_{n=0}^{m}{\frac{\left(-1\right)^{n}}{n!\left(x-n\right)!}} &=\frac{1}{\Gamma\left(1+x\right)}\sum_{n=0}^{m}{\left(-1\right)^{n}\binom{x}{n}}\\ &=\frac{1}{\Gamma\left(1+x\right)}\sum_{n=0}^{m}{\left(\left(-1\right)^{n}\binom{x-1}{n}-\left(-1\right)^{n-1}\binom{x-1}{n-1}\right)}\\ \sum_{n=0}^{m}{\frac{\left(-1\right)^{n}}{n!\left(x-n\right)!}} &=\frac{\left(-1\right)^{m}}{\Gamma\left(1+x\right)}\binom{x-1}{m}\end{aligned}
Taking the limit $ m\longrightarrow +\infty $, we get that : $$ \sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n!\left(x-n\right)!}}=0  $$
A: The binomial coefficient for non-integer $\alpha$ is
$$
\binom{\alpha}{n} = \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}
=\frac{\Gamma(\alpha-1)}{n!\Gamma(\alpha-n+1)}
$$
With the convention $x! = \Gamma(x+1)$ even when $x$ is not a nonnegative integer,
$$
\sum_{n=0}^\infty \frac{(-1)^n}{n!(\alpha-n)!}
=\frac{1}{\Gamma(\alpha+1)}\sum_{n=0}^\infty
 \frac{\Gamma(\alpha+1)(-1)^n}{n!\Gamma(\alpha-n+1)}
= \frac{1}{\Gamma(\alpha+1)}\sum_{n=0}^\infty\binom{\alpha}{n}(-1)^n
$$
In case the series converges, then by Newton's binomial theorem,
$$
\sum_{n=0}^\infty \frac{(-1)^n}{n!(\alpha-n)!} = \frac{(1-1)^\alpha}{\Gamma(\alpha+1)}
$$
The series converges if $\alpha > 0$, of course to the value $0$.  
The series diverges when $\alpha < 0$.  For example, $\alpha=-1/2$:
$$
\sum_{n=0}^\infty\binom{-1/2}{n}(-1)^n
$$
all terms are positive, and the $n$th term is asymptotic to
$$
\frac{1}{\pi\sqrt{n}}
$$
so the series diverges by comparison to $\sum n^{-1/2}$.
