Proving that $0<e-S_{n}<\frac{4}{(n+1)!}$ $\forall n\in\mathbb{N}$

Question

I 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!

Answer 1

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

$$0 < e- S_n < \frac{4}{(n+1)!}$$

Answer 2

Let's consider inequality $$x_n = \left( 1+ \frac{1}{n} \right)^n > 1+ \frac{n}{n} + \frac{n(n-1)}{2!} \frac{1}{n^2} + \cdots + \frac{n(n-1) \dots (n-k+1) }{k!} \frac{1}{n^k}$$ If we now consider limit $n \to \infty$, then we will have $$e \ge 1+1 +\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{k!} = S_k$$ Also we have $$x_n =\left( 1+ \frac{1}{n} \right)^n < 1+1 +\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} = S_n$$ so, there is $$x_n < S_n < e$$ which, after taking limit, finishes proof.

For estimation approximation let's consider $$S_{n+k}-S_n= \frac{1}{(n+1)!}+ \cdots + \frac{1}{(n+k)!}< \frac{1}{(n+1)!}\left( \frac{1}{(n+2)} + \frac{1}{(n+2)^2} +\cdots \right) = \\= \frac{1}{(n+1)!} \frac{n+2}{(n+1)}< \frac{1}{n \cdot n!}$$ Fixing $n$ and taking limit $k \to \infty$ gives $$0<e-S_n<\frac{1}{n \cdot n!}<\frac{4}{(n+1)!}$$