f '(x)=-ln(x+ 1)/(x+ 1)？

F ′ (x) > 0 is -ln(x+ 1)>0, ln (x+1) < 0

Get: 0

∴ The interval of monotonically increasing f (x) is (-1, 0).

The monotone decreasing interval is (0,, +∞)

g(n+ 1)= f[ 1/g(n)- 1]= { 1+ln[ 1/g(n)]}/[ 1/g(n)]

=g(n)[ 1-lng(n)]

Let the function h (x) = x (1-lnx) (0

h'(x)= 1-lnx- 1=-lnx

∫0 & lt； x & lt 1∴h'(x)>； 0

∴h(x) increases monotonously.

∴h(x)<； h( 1)= 1

∵x & gt； 0， 1-lnx & gt； 0∴h(x)>； 0

∴0<； h(x)& lt； 1

∴h(x) When the domain is (0, 1), 0

Inequality 0

1? When n= 1, 0 is known.

,2? Suppose that when n=k, the proposition holds, that is, 0

Then, when n=k+ 1,

g(k+ 1)= g(x)[ 1-g(k)]