How to show that $ \sum_{n = 0}^{\infty} \frac {1}{n!} = e$?

How to show that $\sum\limits_{n = 0}^{\infty} \frac {1}{n!} = e$ where $e = \lim\limits_{n\to\infty} \left({1 + \frac 1 n}\right)^n$?

I'm guessing this can be done using the Squeeze Theorem by applying the AM-GM inequality. But I can only get the lower bound.

If $S_n$ is the $n$th partial sum of our series then,

$$ \left({1 + \frac 1 n}\right)^n = 1 + n\cdot\frac{1}{n} + \frac{n(n - 1)}{2}\cdot \frac{1}{n^2} + \cdots + \frac{n(n - 1)\ldots(n - (n -1))}{n!} \cdot \frac{1}{n^n} $$

$$ \le 1 + \frac{1}{2!} + \cdots + \frac{1}{n!} =S_n $$

How can I show that $S_n \le $ a subsequence of $\left({1 + \frac 1 n}\right)^n$ or any sequence that converges to $e$?

Try bounding

$$E_n = \left ( 1 + \frac{1}{n} \right )^n - \sum_{k=0}^n \frac{1}{k!}$$

using the binomial expansion you wrote, and pairing off terms. That is, write

$$E_n = (1-1) + (1-1) + (1/2 - 1/2n - 1/2) + \dots$$

Using the factorial definition of the binomial coefficient will prove useful.

There is no need to estimate/bound the difference between $\displaystyle\;\left(1 + \frac{1}{n}\right)^n\;$ and $\displaystyle\;\sum_{k=0}^\infty \frac{1}{k!}\;$.

In the expansion

$$\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k} =\sum_{k=0}^\infty \frac{\alpha_{n,k}}{k!}\quad\text{with}\quad\alpha_{n,k} = \begin{cases} \displaystyle\;\prod_{\ell=0}^{k-1}\left(1 -\frac{\ell}{n}\right), & k \le n \\ 0,&k > n \end{cases}$$

If one fix $k$ and look at each $\alpha_{n,k}$ as a sequence of $n$, you find it is non-negative, monotonic increasing and converges to $1$ as $n \to \infty$.

It is clear $\displaystyle\;\sum\limits_{k=0}^\infty \frac{1}{k!}\;$ converges to some finite number. By monotone convergence theorem (for sequences), we can exchange the order of taking limit and summation and get

$$ e \stackrel{def}{=} \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \lim_{n\to\infty}\lim_{N\to\infty} \sum_{k=0}^N \frac{\alpha_{n,k}}{k!} =\lim_{N\to\infty}\lim_{n\to\infty} \sum_{k=0}^N \frac{\alpha_{n,k}}{k!}\\ = \lim_{N\to\infty}\sum_{k=0}^N\frac{1}{k!} \stackrel{def}{=} \sum_{k=0}^\infty\frac{1}{k!} $$ Why does this work? The key is when $\alpha_{n,k}$ is non-decreasing in $n$, we can rewrite the limit as a double sum over non-negative numbers $$\lim_{n\to\infty}\sum_{k=0}^\infty \frac{\alpha_{n,k}}{k!} = \sum_{k=0}^\infty\sum_{\ell=0}^\infty\frac{\alpha_{\ell,k}-\alpha_{\ell-1,k}}{k!} $$ and the sum of any collection of non-negative numbers are independent of the order you perform the summation!

Let $a_n = \left(1+ 1/n\right)^n$. By the binomial theorem,

$$a_n = 1 + 1 + \frac1{2!}\left(1- \frac1{n}\right)+ \ldots +\frac1{n!}\left(1- \frac1{n}\right)\ldots\left(1- \frac{n-1}{n}\right)\leq \sum_{k=0}^{\infty}\frac1{k!}=e,$$

The sequence $a_n$ is increasing and bounded, so it converges.

Hence, $$\lim_{n\rightarrow \infty} a_n \leq e.$$

For $m < n,$

$$a_n \geq 1 + 1 + \frac1{2!}\left(1- \frac1{n}\right)+ \ldots +\frac1{m!}\left(1- \frac1{n}\right)\ldots\left(1- \frac{m-1}{n}\right).$$

Take the limit as $n \rightarrow \infty$ with $m$ fixed to get $$\lim_{n \rightarrow \infty} a_n \geq \sum_{k=0}^{m}\frac1{k!}.$$

Taking the limit again as $m \rightarrow \infty$ we get $$\lim_{n\rightarrow \infty} a_n \geq e.$$

Therefore $$\lim_{n\rightarrow \infty} a_n = e.$$