證明：設(shè)S(n)=1/1²+1/2²+...+1/n²
∵1/1²≤2-1/1
∴猜想S(n)≤2-1/n
當(dāng)n=1時,成立
假設(shè)當(dāng)n=k＞1時成立,即S(k)≤2-1/k
下面正面當(dāng)n=k+1時,S(k+1)≤2-1/(k+1)成立
顯然,S(k+1)=S(k)+1/(k+1)²≤2-1/k+1/(k+1)²
∵k²+2k+1≥k²+2k
(k+1)²≥k(k+1)+k
(k+1)²-k≥k(k+1)
-(k+1)²+k≤-k(k+1)
∵k＞1,∴k+1＞0
-1/k+1/(k+1)²≤-1/(k+1) .//此處不等式左右同時除以k(k+1)²
2-1/k+1/(k+1)²≤2-1/(k+1)
即S(k+1)≤2-1/k+1/(k+1)≤2-1/(k+1),成立
∴猜想成立,即S(n)≤2-1/n
∵1/n＞0
-1/n＜0
2-1/n＜2
∴S(n)≤2-1/n＜2
即S(n)＜2