题目内容
求
（
+
+…+
）．
| lim |
| n→∞ |
| 1 |
| 2! |
| 2 |
| 3! |
| n |
| (n+1)! |
考点：极限及其运算
专题：计算题
分析：由
=
-
，把已知数列的项裂项相消，求和后求极限．
| k |
| (k+1)! |
| 1 |
| k! |
| 1 |
| (k+1)! |
解答：
解：∵
=
-
，
∴
+
+…+
=1-
+
-
+…+
-
=1-
=
．
∴
(
+
+…+
)
=
=1．
| k |
| (k+1)! |
| 1 |
| k! |
| 1 |
| (k+1)! |
∴
| 1 |
| 2! |
| 2 |
| 3! |
| n |
| (n+1)! |
=1-
| 1 |
| 2! |
| 1 |
| 2! |
| 1 |
| 3! |
| 1 |
| n! |
| 1 |
| (n+1)! |
=1-
| 1 |
| (n+1)! |
| (n+1)!-1 |
| (n+1)! |
∴
| lim |
| n→∞ |
| 1 |
| 2! |
| 2 |
| 3! |
| n |
| (n+1)! |
=
| lim |
| n→∞ |
| (n+1)!-1 |
| (n+1)! |
点评：本题考查了极限及其运算，考查了利用裂项相消法求数列的和，是基础题．
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