How do you prove $\sum_{n=0}^\infty \frac{(-1)^n}{n!} = \frac{1}{e}$? prove the sum
$$\sum_{n=0}^\infty \frac{(-1)^n}{n!} = \frac{1}{e}$$
In one of the solutions to a problem I was looking at had this sum and directly got $1/e$ from it. I don't understand how you get that, I used my calculator and it indeed does equal $1/e$ but I'm interested in how you solve this by hand.
 A: Well, given that we know that
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
we can put $x = -1$ to get the series, and the left hand side is then $e^{-1} = \frac{1}{e}$ by exponent rules.
However, perhaps that proof is not entirely satisfying: how do we know the exponent rules work for this series? We may not know yet it is an exponential function or that calling it $e^x$ is justified even if we may have reason to suspect it. Can we prove it without necessarily resorting to them?
Well, going on the supposition it does behave like an exponent, we may be tempted to multiply it with the series for $e$, if we have defined $e$ already by:
$$e = \sum_{n=0}^{\infty} \frac{1}{n!}$$
which gives
$$
\begin{align}
e \times ``\ e^{-1}" &= \left(\sum_{n=0}^{\infty} \frac{1}{n!}\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \right) \\
&= \left(\sum_{n=0}^{\infty} \frac{1}{n!} 1^n\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} 1^n\right)\\
&= \sum_{n=0}^{\infty} \frac{1}{n!} 1^n \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} 1^k\\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{1}{n!} 1^n \frac{(-1)^k}{k!} 1^k\\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{n!k!} 1^{n+k}\\
&= \sum_{l=0}^{\infty} \sum_{n+k=l} \frac{(-1)^k}{n!k!} 1^l\\
&= \sum_{l=0}^{\infty} \left(\sum_{n+k=l} \frac{(-1)^k}{n!k!}\right) 1^l\\
&= \sum_{l=0}^{\infty} \left(\sum_{n+k=l} (-1)^k \frac{1}{l!} \frac{l!}{n!k!}\right) 1^l\\
&= \sum_{l=0}^{\infty} \frac{1}{l!} \left(\sum_{n+k=l} {l \choose n, k} (-1)^k\right) 1^l\\
&= \sum_{l=0}^{\infty} \frac{1}{l!} \left(\sum_{n+k=l} {l \choose n, k} 1^n (-1)^k\right) 1^l\\
&= \sum_{l=0}^{\infty} \frac{1}{l!} (1 + (-1))^l 1^l\\
&= \sum_{l=0}^{\infty} \frac{1}{l!} 0^l 1^l\\
&= 1
\end{align}
$$
where we used the multinomial theorem and that $0^0 = 1$ and $0! = 1$ but $0^l = 0$ for all $l > 0$. Thus the series must sum to the reciprocal of $e$.
A: Using Exponential theorem,we get
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\cdots\infty$
$\Rightarrow e^x=\displaystyle \sum_{n=0}^{\infty}\dfrac{x^n}{n!}$
Now if we put $x=-1$ then we get,$$e^{-1}=\displaystyle\sum_{n=0}^{\infty}\dfrac{{(-1)}^n}{n!}\Rightarrow \dfrac{1}{e}=\displaystyle\sum_{n=0}^{\infty}\dfrac{{(-1)}^n}{n!}$$.
A: You can get it directly from the definition of $e^x$:
$$e^x=\sum_{i=0}^\infty \frac{x^i}{i!}$$
A: Let $$f(x)= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} \dots \infty$$
$$f'(x)=0+\frac{1}{1!}+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!} \dots \infty$$
$$f'(x)= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} \dots \infty$$
$$\implies f'(x)=f(x)$$
Whose general solution is $f(x)=e^x+c$, on comparing the value of $f(0)$, we get $c=0$.
Hence our function is:
$$f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
What we want is ;
$$\sum_{i=0}^\infty \frac{(-1)^n}{n!}=f(-1)=e^{-1}$$
$$ \text{Therefore}~:~~\sum_{n=0}^\infty \frac{(-1)^n}{n!} = \frac{1}{e}$$
A: This is not a new answer, just a more graphical illustration of the answer by @mike4ty4, too long for a comment.
Assuming that all you know is
$$e=\sum_{n=0}^\infty\frac1{n!},$$
you can form the product 
$$\sum_{n=0}^\infty\frac1{n!}\cdot\sum_{n=0}^\infty\frac{(-1)^n}{n!}.$$
We arrange the pairwise products of the terms as a 2D array, of which we show the upper left corner. Notice the alternating signs.
$$\begin{matrix}\frac1{0!0!}&-\frac1{0!1!}&\frac1{0!2!}&-\frac1{0!3!}&\frac1{0!4!}&\cdots\\
\frac1{1!0!}&-\frac1{1!1!}&\frac1{1!2!}&-\frac1{1!3!}&\ddots\\
\frac1{2!0!}&-\frac1{2!1!}&\frac1{2!2!}&\ddots\\
\frac1{3!0!}&-\frac1{3!1!}&\ddots\\
\frac1{4!0!}&\ddots\\
\vdots\\
\end{matrix}$$
Then you can see a connection to Pascal's triangle, by multiplying every diagonal by the successive factorials, which gives integer numbers:
$$\begin{matrix}\frac{0!}{0!0!}&-\frac{1!}{0!1!}&\frac{2!}{0!2!}&-\frac{3!}{0!3!}&\frac{4!}{0!4!}&\cdots\\
\frac{1!}{1!0!}&-\frac{2!}{1!1!}&\frac{3!}{1!2!}&-\frac{4!}{1!3!}&\ddots\\
\frac{2!}{2!0!}&-\frac{3!}{2!1!}&\frac{4!}{2!2!}&\ddots\\
\frac{3!}{3!0!}&-\frac{4!}{3!1!}&\ddots\\
\frac{4!}{4!0!}&\ddots\\
\vdots\\
\end{matrix}$$
$$\begin{matrix}1&-1&1&-1&1&\cdots\\
1&-2&3&-4&\ddots\\
1&-3&6&\ddots\\
1&-4&\ddots\\
1&\ddots\\
\vdots\\
\end{matrix}$$
It turns out that the sum of every diagonal but the first is zero (this is a property of the binomial numbers, verified by expanding $(1-1)^n$), and this holds for the diagonals of the first array.
Hence
$$\sum_{n=0}^\infty\frac1{n!}\cdot\sum_{n=0}^\infty\frac{(-1)^n}{n!}=1.$$
