已知数列{An}的通项公式是An=n/n^2+156(n属于正整数)则数列的最大项是多少

a(n)=n/[n^2 + 156] = 1/[n + 156/n] 1/[n + 156/n] <= 1/[ 2(n*156/n)^(1/2)] = 1/[2(4*39)^(1/2)],什么意思没看懂x为任意正数时，f(x)=1/[x+156/x] <= 1/[2(x*156/x)^(1/2)] = 1/[2(156)^(1/2)],等号成立，当且仅当 x = 156/x, x^2 = 156 = 12*13, 120,b>0时，a+b>=2(a*b)^(1/2). 1/[a+b] <= 1/[2(ab)^(1/2)].当x为正整数n时，a(n) = 1/[n+156/n] <= max[a(12),a(13)],而，a(12)=a(13)=1/25,因此，总有a(n) <= a(12) = a(13) = 1/25.