an=n^(n+1),bn=(n+1)^n 比较大小并证明 用数学归纳法

当n=1时,1^2(k+1)^k,即k^(k+1)/(k+1)^k>1k*(k/(k+1))^k>1当n=k+1时,考察(k+1)^(k+2)>(k+2)^(k+1)是否成立.∵k^2+2k+1>k^2+2k∴(k+1)^2>k(k+2)(k+1)^2/(k+2)>k(k+1)/(k+2)>k/(k+1)((k+1)/(k+2))^k>(k/(k+1))^kk*((k+1...