<s>[INST] Proving that $0I need to prove that $0<e-S_{n}<\frac{4}{(n+1)!}$ $\quad\forall n\in\mathbb{N}\quad$ for $\quad S_{n}=1+1+\frac{1}{2!}+\ldots+\frac{1}{n!}$
Previously, I proved that $2<e<4$, and that $S_n$ is taylor polynomial of $e$.
I thought on proving it using induction:
Base: $S_1=1<2<e<4$
Step - for $n+1$:
$e-S_{n+1}=e-S_{n}-\frac{1}{(n+1)!}$
What am I missing here? It's supposed to be simple.
Thank you!
 [/INST] You have
$$\begin{aligned}0 < e- S_n &= \sum_{k=n+1}^\infty \frac{1}{(k+1)!}\\
&= \frac{1}{(n+1)!}\left(1 + \frac{1}{n+2} + \frac{1}{(n+2)(n+3) }+ \dots\right )\\
&< \frac{1}{(n+1)!}\left(1 + \frac{1}{n+2} + \frac{1}{(n+2)^2 }+ \dots\right )\\
&=  \frac{1}{(n+1)!} \frac{1}{1-\frac{1}{n+2} } = \frac{1}{(n+1)!} \frac{n+2}{n+1} \le \frac{2}{(n+1)!}
\end{aligned}$$
Or using Taylor's theorem:
$$e - S_n = \frac{e^c}{(n+1)!}$$ with $c \in (0,1)$ and with what you already proved
$$0 < e- S_n < \frac{4}{(n+1)!}$$
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