Q:

How to prove an inifnite sum

How can I prove that $$\sum_{n=0}^{\infty}\frac{(-2)^n}{n!}=e^{-2}$$
I do know that $$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^{x}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$$
But I never had a real proof of it and is there an other way to prove the infinite sum above?

A:

Let's prove that series.
$$\lim_{n\to \infty}\left(1 + \frac{x}{n}\right)^n$$
From the binomial theorem we have
$$
\begin{align}
\left(1 + \frac{x}{n}\right)^n & = 1 + x + \frac{n(n-1)}{2!n^2}x^2 + \frac{n(n-1)(n-2)}{3! n^3}x^3 + \cdots
\\\\
& = \frac{x^0}{0!} + \frac{x^1}{1!} + \left(\frac{n-1}{n}\right)\frac{x^2}{2!} + \left(\frac{(n-1)(n-2)}{n^2}\right)\frac{x^3}{3!} + \cdots 
\end{align}
$$
Now as $n\to \infty$ each term in the brackets is $1$
$$\lim_{n\to \infty}  \left(\frac{n-1}{n}\right) = 1$$
$$\lim_{n\to \infty}  \left(\frac{(n-1)(n-2)}{n^2}\right) = 1$$
Et cetera.
What remains is
$$
\begin{align}
\lim_{n\to \infty}\left(1 + \frac{x}{n}\right)^n & = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
\\\\
& = \sum_{k = 0}^{+\infty} \frac{x^k}{k!}
\\\\
& = e^x
\end{align}
$$
Thence use $x = -2$ and make a backwards calculation, et voilà.

