Doubt in a step.

\left(1+\frac{1}{n} \right)^n = 1 + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n^2}+....

= 2+ \frac{1}{2} \left(1-\frac{1}{n} \right)\left(1-\frac{2}{n} \right)+\frac{1}{3!} \left(1-\frac{1}{n} \right)\left(1-\frac{2}{n} \right) \left(1-\frac{3}{n} \right)

< 2+ \frac{1}{2!} +\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}

But we know that n!> 2^n for n>3.

So we continue the chain of inequalities by noting that the quantity is therefore less than 2 + \frac{1}{2} + \frac{1}{4}+...<3

@theprophet

Thank you for your reply.
Is this fine?-

\textup{Using}\ \frac{1}{n!}< \frac{1}{2^{n}}\ \textup{for}\ n> 3 \\\ 2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+.....< 2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{2^{4}}+\frac{1}{2^{5}}+....< 3