用数学归纳法最简单：
证明：当 n = 1 时,1+n/2=3/2 ,1+1/2+1/3+……+1/2^n=3/2 ,1/2+n=3/2 ,显然不等式成立.
令 n = k 时,不等式成立
即,1+k/2 ≤ 1+1/2+1/3+……+1/2^k ≤ 1/2+k
令 f(k) = 1+1/2+1/3+……+1/2^k
有 f(k+1) = f(k)+1/(2^k+1)+1/(2^k+2)+……+1/2^(k+1)
∵ 1/(2^k+1) ≥ 1/2^(k+1) ,1/(2^k+2) ≥ 1/2^(k+1) ,……,1/2^(k+1) ≥ 1/2^(k+1)
∴ f(k+1) ≥ f(k)+1/2^(k+1)+1/2^(k+1)+.+1/2^(k+1) [共2^k个]
∴ f(k+1) ≥ f(k)+[1/2^(k+1)] * 2^k = f(k) + 1/2 ≥ 1+k/2 + 1/2 = 1+ (k+1)/2
同理 ∵ 1/(2^k+1) ≤ 1/2^k ,1/(2^k+2) ≤ 1/2^k ,…… ,1/2^(k+1) ≤ 1/2^k
∴ f(k+1) ≤ f(k)+ (1/2^k) * 2^k = f(k) +1 ≤ 1/2 + (k + 1)
所以 1+ (k+1)/2 ≤ f(k+1) = [1+1/2+1/3+……+1/2^(k+1) ] ≤ 1/2 + (k + 1)
综上 (1+n/2)≤1+1/2+1/3+……+1/2^n≤（1/2+n）对于n≥1均成立.