设n为正整数,讨论函数f(n)=(n+1)*(0.9)

f(n)>0
作商比较
f(n+1)/f(n)
=[(n+2)*(0.9)^(n+1)]/[(n+1)*(0.9)^n]
=0.9*(n+2)/(n+1)

当f(n+1)/f(n)>1
0.9*(n+2)/(n+1)>1
解得n<8
即f(1)<f(2)<...<f(8)

当当f(n+1)/f(n)=1
0.9*(n+2)/(n+1)=1
解得n=8
即f(8)=f(9)

当f(n+1)/f(n)<1
0.9*(n+2)/(n+1)<1
解得n>8
即f(9)>f(10)>...>f(n)

综上所述f(1)<f(2)<...<f(8)=f(9)>f(10)>...
f(n)取最大值 n=8或9
f(8)=f(9)=9*0.9^8

f(n+1)/f(n) = 0.9*(n+2)/(n+1)>=1
=>
n<=8
n=8时，f(8)=f(9）
=>
f(n)最大
n = 8或9