Question

高中数学：数列与不等式

求证：n＜(k=1，n)∑√(1+(1/k²)+(1/(k+1)²))＜n+1. (n∈N+)

Answer 1

1+ 1/k²+ 1/(k+1)²
=[k²(k+1)²+(k+1)²+k²]/[k²(k+1)²]
=[k²(k+1)²+2k²+2k+1]/[k²(k+1)²]
=[k²(k+1)²+2k(k+1)+1]/[k²(k+1)²]
=[k(k+1)+1]²/[k(k+1)]²
√[1+ 1/k²+ 1/(k+1)²]
=√[k(k+1)+1]²/[k(k+1)]²
=[k(k+1)+1]/[k(k+1)]
=1+ 1/[k(k+1)]
=1+ 1/k -1/(k+1) /关键就是化简，剩下的就非常简单了。

n
∑ √[1+ 1/k²+ 1/(k+1)²]
k=1
=1+1-1/2+1+1/2-1/3+...+1+1/n -1/(n+1)
=n+[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=n+1 -1/(n+1)
1/(n+1)>0 n+1- 1/(n+1)<n+1
n
∑ √[1+ 1/k²+ 1/(k+1)²] <n+1
k=1

n≥1 1/(n+1)≤1/2
n+1 -1/(n+1)≥n+1/2>n

n
∑ √[1+ 1/k²+ 1/(k+1)²]>n
k=1

综上，得

n
n< ∑ √[1+ 1/k²+ 1/(k+1)²] <n+1

提示：本题的关键是化简去根号，后面就非常简单了。

Answer 2