Belarusian Mathematical Olympiad

Transform the expression on the left side $$\begin{array}{rl}\frac{1}{2!}+\frac{2}{3!}+\cdots +\frac{2^{n-2}}{n!}& =\frac{1}{2!}\left(1+\frac{2}{3}+\frac{2^2}{3\cdot 4}+\cdots +\frac{2^{n-2}}{3\cdot 4\dots n}\right)\\ & \le \frac{1}{2}\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots +\frac{2^{n-2}}{3^{n-2}}\right)\\ & =\frac{1}{2}\left(1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+\cdots +{\left(\frac{2}{3}\right)}^{n-2}\right)\\ & \le \frac{1}{2}\left(1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+\cdots +{\left(\frac{2}{3}\right)}^{n-2}+\dots \right)\\ & =\frac{1}{2}\cdot \frac{1}{1-\frac{2}{3}}=\frac{3}{2}.\end{array}$$

Second solution:

We will prove by induction a stronger inequality $$\frac{1}{2!}+\frac{2}{3!}+\cdots +\frac{2^{n-2}}{n!}\le \frac{3}{2}-\frac{1}{n},(\ast )$$ from which, obviously, follows the required. It is easy to check that at $n=2,3,4,5$ the inequality $(\ast )$ holds — it is the base of induction.

Induction step: assume that the inequality $(\ast )$ is true for some $n=k$. To prove $(\ast )$ for $n=k+1$ it is enough to show that the right side of this inequality increases faster than the left side, i.e. it is enough to prove the inequality $$\frac{2^{k-1}}{(k+1)!}\le \frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}.$$