求证:1/n+1+1/n+2+…+1/3n>5/6(n≥2,n∈N*).

问题描述:

求证:

1
n+1
+
1
n+2
+…+
1
3n
5
6
(n≥2,n∈N*).

证明:(1)当n=2时,左边=

1
3
+
1
4
+
1
5
+
1
6
57
60
50
60
5
6
,不等式成立;
(2)假设n=k(k≥2,k∈N*)时命题成立,即
1
k+1
+
1
k+2
+…+
1
3k
5
6
成立.
则当n=k+1时,左边=
1
(k+1)+1
+
1
(k+1)+2
+
…+
1
3k
+
1
3k+1
+
1
3k+2
+
1
3(k+1)

=
1
k+1
+
1
k+2
+
…+
1
3k
+
(
1
3k+1
+
1
3k+2
+
1
3k+3
1
k+1
)

5
6
+
(3×
1
3k+3
1
k+1
)
=
5
6

所以当n=k+1时不等式也成立.
综上由(1)(2)可知:原不等式对任意n≥2(n∈N*)都成立.