这种类型的题通常用取对数来降次证明：∵k^(k+1)>（k+1）^k∴(k+1)lnk>kln(k+1)∴k[lnk-ln(k+1)]+lnk>0∴kln[k/(k+1)]+lnk>0 .(1)又(k+1)^2=k+2k+1>k+2k=k(k+2)∴(k+1)/(k+2)>k/(k+1)∴ln[(k+1)/(k+2)]>ln[k/(k+1)]....