I was playing around with factorials and integrals when I put $ \sum_{x=1}^ \infty \frac{1}{x!} $ in Wolfram Alpha
I got the result as equal to $e - 1$.
It was an amazing discovery for me.
I tried to prove it, but could go nowhere. I just know that $\lim_{x \rightarrow \infty} \big(1 + \frac{1}{x} \big)^{x} = e$.
Can anyone provide a pointer to what should I do ?
EDIT : Everyone is referring to the definition of the exponential function. Can you provide a more elementary proof ?
Intuitively, when $x$ is a whole number, we can take the binomial expansion:
$$\left(1+\frac1x\right)^x=1+1+\frac12\frac{x-1}{x}+\frac16\frac{(x-1)(x-2)}{x^2}+\dots$$
And as $x\to\infty$, each fraction goes to $1$, so we have
$$e=\lim_{x\to\infty}\left(1+\frac1x\right)^x=1+\sum_{n=1}^\infty\frac1{n!}$$
Though you need to make this more rigorous at the limit part if you want a good proof.
Let $e$ be the number represented as the limit
$$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n$$
It is easy to show (See Here) that the sequence $\left(1+\frac1n\right)^n$ is monotonically increasing. It is also straightforward to show that the sequence is bounded above by $3$. Hence, the limit exists.
From the binomial theorem, we can write
$$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n=\lim_{n\to \infty}\sum_{k=0}^n\binom{n}{k}\frac1{n^n}$$
Next, we note that for all $0\le k\le n$
$$\binom{n}{k}\frac{1}{n^n}=\frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right)\le \frac1{k!}$$
Therefore, we see that
$$\left(1+\frac1n\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^n}\le \sum_{k=0}^n\frac1{k!}$$
Taking the $\limsup$ as $n\to \infty$ reveals
$$\begin{align} e&\le \limsup_{n\to \infty}\sum_{k=0}^n\frac1{k!}\\\\ &=\lim_{n\to \infty}\sum_{k=0}^n\frac1{k!}\tag1 \end{align}$$
where we used the fact that $\sum_{k=0}^\infty\frac1{k!}$ converges (apply the ratio test, for example) to arrive at $(1)$.
Next, we fix $m\le n$. Then, clearly
$$\begin{align} \left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^n}\\\\ &\ge \sum_{k=0}^m \left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right)\frac1{k!} \end{align}$$
With $m$ fixed, we take the limit as $n\to \infty$ and find that
$$e\ge \sum_{k=0}^m\frac{1}{k!} \tag 2$$
Then, we let $m\to \infty$ in $(2)$ to obtain
$$e\ge \sum_{k=0}^\infty\frac1{k!} \tag 3$$
Finally, putting together $(1)$ and $(3)$ yields
$$e=\sum_{k=0}^\infty\frac{1}{k!}$$
whence we see that
$$e-1=\sum_{k=1}^\infty \frac{1}{k!}$$
And we are done!
