| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 13 |
| 12 |
（2）假设当n=k（k≥2）时成立，即
| 1 |
| k |
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| k2 |
那么当n=k+1时，左边=
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| (k+1)2 |
=
| 1 |
| k |
| 1 |
| k+1 |
| 1 |
| k+2 |
| 1 |
| k+3 |
| 1 |
| k2+2k |
| 1 |
| (k+1)2 |
| 1 |
| k |
＞1+
| 1 |
| k2+1 |
| 1 |
| k2+2 |
| 1 |
| (k+1)2 |
| 1 |
| k |
＞1+(2k+1)•
| 1 |
| (k+1)2 |
| 1 |
| k |
| k2−k−1 |
| k2+2k+1 |
∴n=k+1时也成立（7分）
根据（1）（2）可得不等式对所有的n＞1都成立（8分）