Belarusian Mathematical Olympiad

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

Second solution:

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

Induction step: assume that the inequality $(*)$ is true for some $n = k$. To prove $(*)$ 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
$$\frac{2^{k-1}}{(k+1)!} \le \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}.$$