<s>[INST] Help in proving $ \left( 1 + \frac{1}{n} \right)^{n} \leq \sum\limits_{k=0}^{n} \frac{1}{k!} < 3 $. I am trying to prove this statement for all $ n \geq 1 $ using induction:
$$
\left( 1 + \frac{1}{n} \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3.
$$
I said:


*

*Base case $ n = 1 $:
$$
  \left( 1 + \frac{1}{1} \right)^{1} \leq \sum_{k=0}^{1} \frac{1}{k!} < 3,
  $$
which is okay.

*Induction step: Suppose that $ \displaystyle \left( 1 + \frac{1}{n}
  \right)^{n} \leq \sum_{k=0}^{n} \frac{1}{k!} < 3 $ for a given $ n \in
  \mathbb{N} $.
Transition from $ n \to n + 1 $:
$$
    \displaystyle \left( 1 + \frac{1}{n + 1} \right)^{n+1}
  = \left( 1 + \frac{1}{n + 1} \right)^{n} \left( 1 + \frac{1}{n + 1} \right)
  = \ldots \text{Help} \ldots
  < 3.
  $$
I need some guidance for proof-writing (-thinking) in orders.
 [/INST] To show that $\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!} \lt 3$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots +\frac{1}{n!}$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}=1+1+\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{1}{3*4}+\frac{1}{3*4*5}+\dots+\frac{1}{3*4*5*\dots*n}\right)$
$\displaystyle\sum_{k=0}^{k=n}\frac{1}{k!}\lt1+1+1$
</s>