determining the value of e by using mathematical induction

Question

using the fact that $n!>2^{n}$ $\forall n\ge 4$ conclude that $e<\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\displaystyle\sum_{n=4}^{\infty}\frac{1}{2^{n}}$

where $e:=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}=1+\frac{1}{1!}+....$

should I inverse the inequality

$\frac{1}{n!}<\frac{1}{2^{n}}$ $\forall n\ge 4$ ??

Please help

Answer 1

Write $e=1+1/1!+1/2!+1/3!+\sum_{k=4}^\infty 1/n!$ and use your inequality for the summands of the summation.

Answer 2

If you start from $n! \gt 2^n$ and divide both sides by $2^nn!$ you get the inequality you want.