Prove the Inequality on sequence

Question

$a_n=(1+\frac{1}{n})^n$ ,

$b_n=\sum_{k=0}^n \frac{1}{k!}$.

Show that $b_n-\frac{3}{2n} < a_n < b_n$.

Answer 1

From the binomial theorem we have

$$\begin{align} \left(1+\frac1n \right)^n &=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\ &=\sum_{k=0}^n \frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}\frac{1}{k!}\\\\ &=\sum_{k=0}^n \left(\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\right)\frac{1}{k!}\\\\ &\le \sum_{k=0}^n\frac{1}{k!}\\\\ \end{align}$$

since all of the terms $\left(1-\frac{j}{n}\right)<1$.