Răspuns:
Avem
[tex]\displaystyle\frac{1}{k(k+1)(k+2)}=\frac{1}{2}\left[\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right][/tex]
Atunci
[tex]\displaystyle\sum_{k=1}^n\frac{1}{k(k+1)(k+2)}=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{(n+1)(n+2)}\right]=\frac{n^2+3n}{4(n^2+3n+2)} < \frac{1}{4}[/tex]
Explicație pas cu pas:
