Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$ Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$
I proved it by induction:
"$n=1$"   $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad  \checkmark$
"$n \implies n+1$": $$\quad \sum_{k=0}^{n+1}{\frac{1}{k!}}\leq 3 \\ \left(\sum_{k=0}^{n}{\frac{1}{k!}}\right)+\frac{1}{(n+1)!}\leq 3 \\ $$$$3+\overbrace{\frac{1}{(n+1)!}}^{\geq 0}\leq 3 \\$$ That's of course true.

For any finite $n$, 
$$\sum_{k=0}^n \frac{1}{k!} < e < 3.$$
UPDATE
$$1+1+\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n! }< 1 + 1+ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} = 1+2 - \frac{1}{2^{n-1}} <3.$$
