用数学归纳法
证明
当n=1时,
左边=1=右边
当n=2时
左边=1+1/4=5/4
右边=6/5
左边>右边
假设当n=k,k∈N时成立,则
1/1+1/(2^2)+1/(3^2)+1/(4^2)+……+1/(k^2)≥3k/(2k+1)
当n=k+1时
1/1+1/(2^2)+1/(3^2)+1/(4^2)+……+1/(k^2)+1/(k+1)^2
≥3k/(2k+1)+1/(k+1)^2
1/(k+1)^2-3/(2k+1)(2k+3)
=[(2k+1)(2k+3)-3(k+1)^2]/(k+1)^2(2k+1)(2k+3)
=(k^2+2k)/(k+1)^2(2k+1)(2k+3)
因为k为正,所以上式大于0
1/(k+1)^2>3/(2k+1)(2k+3)
原式>3k/(2k+1)+3/(2k+1)(2k+3)
=[3k*(2k+3)+3]/(2k+1)(2k+3)
=3(2k²+3k+1)/(2k+1)(2k+3)
=3(2k+1)(k+1)/(2k+1)(2k+3)
=3(k+1)/(2k+3)
=3(k+1)/[2(k+1)+1]
也成立
所以1/1+1/(2^2)+1/(3^2)+1/(4^2)+……+1/(n^2)≥3n/(2n+1)