Ta có: k2 > k2 - 1 = (k-1)(k+1)
⇒ 1/k2 < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*)
Áp dụng (*), ta có:
1/22 + 1/32 + 1/42 + ... + 1/n2
< 1/22 + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)]
= 1/22 + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2
= 1/22 + [1/2 + 1/3 - 1/n - 1/(n+1)]/2
= 2/3 - [1/n + 1/(n+1)]/2 < 2/3 < 1