题目内容
| lim |
| n→∞ |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
∵
+
+…+
=
(
+
+…+
)
∴
（
+
+…+
）
=
[
(
+
+…+
)]
令x=
，则n→∞，
→0，
=1
则
[
(
+
+…+
)]
=
dx=
=ln2
故答案为：ln2
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
=
| 1 |
| n |
| 1 | ||
1+
|
| 1 | ||
1+
|
| 1 | ||
1+
|
∴
| lim |
| n→∞ |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
=
| lim |
| n→∞ |
| 1 |
| n |
| 1 | ||
1+
|
| 1 | ||
1+
|
| 1 | ||
1+
|
令x=
| 1 |
| n |
| 1 |
| n |
| n |
| n |
则
| lim |
| n→∞ |
| 1 |
| n |
| 1 | ||
1+
|
| 1 | ||
1+
|
| 1 | ||
1+
|
=
| ∫ | 10 |
| 1 |
| 1+x |
| ln(1+x)| | 10 |
故答案为：ln2
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相关题目
| lim |
| n→∞ |
| 1 |
| n+1 |
| 2 |
| n+1 |
| 3 |
| n+1 |
| 2n-1 |
| n+1 |
| 2n |
| n+1 |
| A、-1 | ||
| B、0 | ||
C、
| ||
| D、1 |