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