Prove inequality by induction $\forall n \in \mathbb{N}, n\ge 2 \ \ : \ \ (1+\frac1n)^n<\sum\limits_{k=0}^n \frac1{k!}<3$

Question

I have the following statement to prove: $$\forall n \in \mathbb{N}, n\ge 2 \ \ : \ \ (1+\frac1n)^n<\sum_{k=0}^n \frac1{k!}<3$$

What I have already proven is that $$(1+\frac1n)^n< (1+\frac1{n+1})^{n+1}$$

I have started, as usual by induction with $n = 2$. Then I went on to say $$(1+\frac1{n+1})* (1+\frac1{n+1})^{n}< \sum_{k=0}^n \frac1{k!} + \frac1{(n+1)!} < 3$$

But this is where I seem to get stuck. I cannot use the induction hypothesis, since the denominator on the left side is now $n+1$ instead of $n$. I assume, I have to use the other inequality, which I have already proven to be true, but I do not know how. Any hints?

Answer 1

By Binom Newton $$\left(1+\frac{1}{n}\right)^n=2+\frac{1-\frac{1}{n}}{2!}+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}{3!}+...+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)}{n!}<$$ $$<2+\frac{1}{2!}+\frac{1}{3!}...+\frac{1}{n!}<2+\frac{1}{2}+\frac{1}{2^2}+...=3$$