Question 100
Prove that for every positive integer $n$ the inequality
$$
\frac{3}{1!+ 2!+ 3!} + \frac{4}{2!+ 3!+ 4!} + \dots + \frac{n+2}{n!+ (n+1)!+ (n+2)!} \lt \frac{1}{2}
$$
holds. (Here $1!=1$, $k!=1 \cdot 2 \cdot \dots \cdot (k-1) \cdot k$, $k \ge 2$.)
$$
\frac{3}{1!+ 2!+ 3!} + \frac{4}{2!+ 3!+ 4!} + \dots + \frac{n+2}{n!+ (n+1)!+ (n+2)!} \lt \frac{1}{2}
$$
holds. (Here $1!=1$, $k!=1 \cdot 2 \cdot \dots \cdot (k-1) \cdot k$, $k \ge 2$.)