证明ln2/2^2+ln3/3^2+.+lnn/n^2
问题描述:
证明ln2/2^2+ln3/3^2+.+lnn/n^2
数学人气:832 ℃时间:2020-02-03 07:01:17
优质解答
当n=2时,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;假设当n=k(k≥2为正整数)时不等式成立,即ln2/2^2+ln3/3^2+...+lnk/k^2<(2k^2-k-1)/[4(k+1)]成立,在此不等式两端同时加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,容易证明ln(k+1)<k+1,所以①式变为ln2/2^2+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,容易证明(2k^2-k+3)/(4k+4)<[2(k+1)^2-(k+1)-1]/4[(k+1)+1]③,该不等式可化简得3<k(k+1)④,由于k≥2,所以④成立,进而③成立,所以②式变为ln2/2^2+ln3/3^2+...+lnk/k^2+ln(k+1)/(k+1)^2<[2(k+1)^2-(k+1)-1]/4[(k+1)+1],即当n=k+1时,不等式也成立;所以对所有n≥2的正整数,原不等式均成立.
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答
当n=2时,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;假设当n=k(k≥2为正整数)时不等式成立,即ln2/2^2+ln3/3^2+...+lnk/k^2<(2k^2-k-1)/[4(k+1)]成立,在此不等式两端同时加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,容易证明ln(k+1)<k+1,所以①式变为ln2/2^2+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,容易证明(2k^2-k+3)/(4k+4)<[2(k+1)^2-(k+1)-1]/4[(k+1)+1]③,该不等式可化简得3<k(k+1)④,由于k≥2,所以④成立,进而③成立,所以②式变为ln2/2^2+ln3/3^2+...+lnk/k^2+ln(k+1)/(k+1)^2<[2(k+1)^2-(k+1)-1]/4[(k+1)+1],即当n=k+1时,不等式也成立;所以对所有n≥2的正整数,原不等式均成立.