<s>[INST] Infinite series for $e$... How do you prove that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$? Here I am assuming $e:=\lim_{n\to\infty}(1+\frac{1}{n})^n$. Do you have any good PDF file or booklet available online on this? I do not like how my analysis text handles this...
 [/INST] First prove by the ratio test that the series $\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!}$ converges and denote the sum by $S$.
Then note that
$$\begin{align*}
\left( 1 + \frac{1}{n} \right)^n & = \sum_{k=0}^n \binom{n}{k} \cdot 1^{n-k} \cdot \left( \frac{1}{n} \right)^k = \sum_{k=0}^n \frac{n (n-1) \ldots (n-k+1)}{k!} \cdot \frac{1}{n^k} \\[1ex]
& = \sum_{k=0}^n \frac{1}{k!} \cdot \left( 1 - \frac{1}{n} \right)\left(1-\frac{2}{n}\right) \ldots \left( 1 - \frac{k-1}{n} \right) = \sum_{k=0}^n \frac{1}{k!} \cdot P_n^{(k)}
\end{align*}$$
where
$$P_n^{(k)} = \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \ldots \left(1-\frac{k-1}{n}\right).$$
We can see that for each $k \in \mathbb{N}$ we have $0 \leqslant P_n^{(k)} \leqslant 1$ and $\displaystyle \lim_{n \to \infty} P_n^{(k)} = 1$. 
Fix $\varepsilon > 0$ and let $K$ be so large that $\displaystyle \sum_{k=0}^K \frac{1}{k!} \geqslant (1-\varepsilon)S$. Now let $N$ be so large that whenever $n \geqslant N$, for $k = 0, 1, \ldots, K$ we have $P_n^{(k)} > 1-\varepsilon$.
So for $n \geqslant N$
$$\begin{align*}
(1-\varepsilon)^2 S \leqslant (1-\varepsilon) \sum_{k=0}^K \frac{1}{k!} \leqslant \sum_{k=0}^K \frac{1}{k!} P_n^{(k)} \leqslant \left(1+\frac{1}{n}\right)^n \leqslant \sum_{k=0}^n \frac{1}{k!} \leqslant S.
\end{align*}$$
Therefore 
$$e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = S = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
</s>