Then f (m)-f (n) = am+(m-2)/(m+1)-an n-(n-2)/(n+1).

=(a^m-a^n)+[(m-2)(n+ 1)-(n-2)(m+ 1)]/(m+ 1)(n+ 1)

=(a^m-a^n)+3(m-n)/(m+ 1)(n+ 1)

a & gt 1，m & ltn

So an m

m & ltn，m-n & lt； 0

m & gt- 1，n & gt- 1，m+ 1 & gt； 0，n+ 1 & gt； 0

So (m+ 1)(n+ 1)>0.

so 3(m-n)/(m+ 1)(n+ 1)< 0

Therefore, (a m-a n)+3 (m n)/(m+1) (n+1) < 0.

That is-1< m <; N times

f(m)& lt； coedna

So f(x) is increasing function at (-1, +∞).