當(dāng)n=2時(shí),不等式左端=ln2/2^2,不等式右端=5/12 ,ln2/2^2＜5/12,不等式成立；假設(shè)當(dāng)n=k(k≥2為正整數(shù))時(shí)不等式成立,即ln2/2^2＋ln3/3^2＋...＋lnk/k^2＜(2k^2-k-1)/[4(k＋1)]成立,在此不等式兩端同時(shí)加ln(k＋1)/(k＋1)^2得ln2/2^2＋ln3/3^2＋...＋lnk/k^2＋ln(k＋1)/(k＋1)^2＜(2k^2-k-1)/(4k＋4)＋ln(k＋1)/(k＋1)^2①,容易證明ln(k＋1)＜k＋1,所以①式變?yōu)閘n2/2^2＋ln3/3^2＋...＋lnk/k^2＋ln(k＋1)/(k＋1)^2＜(2k^2-k-1)/(4k＋4)＋(k＋1)/(k＋1)^2=(2k^2-k＋3)/(4k＋4)②,容易證明(2k^2-k＋3)/(4k＋4)＜[2(k＋1)^2-(k＋1)-1]/4[(k＋1)＋1]③,該不等式可化簡(jiǎn)得3＜k(k＋1)④,由于k≥2,所以④成立,進(jìn)而③成立,所以②式變?yōu)閘n2/2^2＋ln3/3^2＋...＋lnk/k^2＋ln(k＋1)/(k＋1)^2＜[2(k＋1)^2-(k＋1)-1]/4[(k＋1)＋1],即當(dāng)n=k＋1時(shí),不等式也成立；所以對(duì)所有n≥2的正整數(shù),原不等式均成立.