(1)当n∈N+时,求证:12≤1n+1+1n+2+…+12n<1;(2)当n∈N+时,求证:1+122+132+…+1n2<2.-高二数学

题目简介

(1)当n∈N+时,求证:12≤1n+1+1n+2+…+12n<1;(2)当n∈N+时,求证:1+122+132+…+1n2<2.-高二数学

题目详情

(1)当n∈N+时,求证:
1
2
1
n+1
+
1
n+2
+…+
1
2n
<1;
(2)当n∈N+时,求证:1+
1
22
+
1
32
+…+
1
n2
<2.
题型:解答题难度:中档来源:不详

答案

(1)证明:∵class="stub"1
2n
+class="stub"1
2n
+class="stub"1
2n
+…+class="stub"1
2n
class="stub"1
n+1
+class="stub"1
n+2
+…+class="stub"1
2n
class="stub"1
n
+class="stub"1
n
+…+class="stub"1
n

class="stub"1
2
class="stub"1
n+1
+class="stub"1
n+2
+…+class="stub"1
2n
<1,故不等式成立.
(2)证明:∵1+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
<1+class="stub"1
1×2
+class="stub"1
2×3
+class="stub"1
3×4
+…+class="stub"1
(n-1)×n

=1+1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n-1
-class="stub"1
n
=2-class="stub"1
n
<2,
即 1+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
<2.

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