题目内容
（1）当n∈N+时，求证：| 1 |
| 2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
（2）当n∈N+时，求证：1+
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
分析：（1）利用
+
+
+…+
≤
+
+…+
＜
+
+…+
，进行放缩．
（2）利用 1+
+
+…+
＜1+
+
+
+…+
=1+1-
+
-
+
-
+…+
-
=2-
，得到要证的结果．
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
| 1 |
| n |
| 1 |
| n |
| 1 |
| n |
（2）利用 1+
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| (n-1)×n |
=1+1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n-1 |
| 1 |
| n |
| 1 |
| n |
解答：解：（1）证明：∵
+
+
+…+
≤
+
+…+
＜
+
+…+
，
∴
≤
+
+…+
＜1，故不等式成立．
（2）证明：∵1+
+
+…+
＜1+
+
+
+…+
=1+1-
+
-
+
-
+…+
-
=2-
＜2，
即 1+
+
+…+
＜2．
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
| 1 |
| n |
| 1 |
| n |
| 1 |
| n |
∴
| 1 |
| 2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
（2）证明：∵1+
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| (n-1)×n |
=1+1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n-1 |
| 1 |
| n |
| 1 |
| n |
即 1+
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
点评：本题考查用放缩法证明不等式，掌握好放缩的程度，是解题的难点．
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