<s>[INST] 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.
 [/INST] $$\sum_{k=0}^{\infty}\frac{1}{k!} = 1 + 1 + \frac{1}{2!}+\frac{1}{3!}+ \cdots$$ $$ < 1+1+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5}+ \cdots$$ $$ = 1 + 1 +\left (1-\frac{1}{2}\right) + \left (\frac{1}{2}-\frac{1}{3}\right) + \left (\frac{1}{3}-\frac{1}{4}\right)+\cdots$$ $$ = 1+1 +1 =3.$$
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