Prove that the sum of inverses of factorials is convergent Let $S_n = \sum\limits_{k=0}^n \dfrac{1}{k!}$.
I must show that the sequence $S_n$ is a Cauchy Sequence, which means:
$$\forall\epsilon\gt 0 \text{ }\text{ }\text{ }\text{ } \exists n_0 \in \mathbb{N} \text{ }\text{ }\text{ }\text{ } \forall n,m \in \mathbb{N} \text{ }\text{ }\text{ }\text{ } (n \gt n_0 \text{ } \land \text{ } m \gt n_0 \implies |S_m - S_n| < \epsilon) $$
This is what I've tried so far:
Assuming WLOG $m > n$
$$|S_m - S_n| = \sum\limits_{k=n+1}^m \dfrac{1}{k!} = \dfrac{1}{m!} \sum\limits_{k=n+1}^m \prod\limits_{i=k+1}^m i \leq \dfrac{1}{m!} \sum\limits_{k=n+1}^m \prod\limits_{i=k+1}^m m \leq \dfrac{1}{m!} \sum\limits_{k=n+1}^m m^{m-k}$$
$$\implies |S_m - S_n| \leq \dfrac{m^m}{m!}\sum\limits_{k=n+1}^m \dfrac{1}{m^k} = \dfrac{m^{m-n}-1}{m!(m-1)}$$
But now I am stuck. I realize my attempt is probably going anywhere because of the substitution that $i \leq m$, which is a big change. Still, I decided to see if I could get somewhere, but got stuck.
I have two questions.
1. Is it possible to continue my approach and complete the proof?
2. What would be a better way to handle this?
Note: I know this sequence converges to $e$ because this is exactly the Taylor Series for the function $e^x$ when $x = 0$, but I don't want to use that fact. I would like to proceed as the question asks, and prove that it is a Cauchy Sequence first, to only then conclude that it is convergent.
Note 2: I would tag this as homework-and-exercises but I couldn't find that tag, maybe I saw this tag in another SE instead of Math.SE.
 A: $$\begin{align}
\sum_{k=0}^n \frac{1}{k!}&=2+\sum_{k=2}^n \frac{1}{k!}\\
&\leq 2+\sum_{k=2}^n\frac{1}{k(k-1)}\\
&=2+\sum_{k=2}^n \left(\frac{1}{k-1}-\frac{1}{k}\right)\\
&=2+\sum_{k=2}^n \frac{1}{k-1}-\sum_{k=2}^n\frac{1}{k}\\
&=2+1-\frac{1}{n}\\
\end{align}
$$
The series is monotonous because $\frac{1}{k!}\geq 0$ and has an upper limit the series converges.
A: When I have shown this,
I base my bounds 
on the lower index,
not the upper.
$\begin{array}\\
|S_m - S_n| 
&= \sum\limits_{k=n+1}^m \dfrac{1}{k!} \\
&= \dfrac{1}{n!}\sum\limits_{k=n+1}^m \dfrac{n!}{k!} \\
&= \dfrac{1}{n!}\sum\limits_{k=n+1}^m \dfrac{1}{\prod_{j=n+1}^k j} \\
&\le \dfrac{1}{n!}\sum\limits_{k=n+1}^m \dfrac{1}{\prod_{j=n+1}^k (n+1)} \\
&= \dfrac{1}{n!}\sum\limits_{k=n+1}^m \dfrac{1}{(n+1)^{k-n}} \\
&= \dfrac{1}{n!}\sum\limits_{k=1}^{m-n} \dfrac{1}{(n+1)^{k}} \\
&< \dfrac{1}{n!}\sum\limits_{k=1}^{\infty} \dfrac{1}{(n+1)^{k}} \\
&= \dfrac{1}{n!}\frac{1/(n+1)}{1-1/(n+1)} \\
&= \dfrac{1}{n\cdot n!} \\
\end{array}
$
and this goes to zero
as $n \to \infty$.
