用数学归纳法证明：1/1^2+1/2^2+.+1/n^2

证明：设S(n)=1/1²+1/2²+。。。+1/n²∵1/1²≤2-1/1∴猜想S(n)≤2-1/n当n=1时,成立假设当n=k＞1时成立,即S(k)≤2-1/k下面正面当n=k+1时,S(k+1)≤2-1/(k+1)成立显然,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) 。//此处不等式左右同时除以k(k+1)²2-1/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＜02-1/n＜2∴S(n)≤2-1/n＜2即S(n)＜2