1

n

xn+

1

nitrogen

-2

∴f′(x)=

1

x

-xn- 1，

Let f'(x)=0, then x= 1,

When x ∈ (0, 1), f ′ (x) > 0 and the function is increasing function.

When x∈( 1, +∞) and f ′ (x) < 0, the function is a decreasing function.

Therefore, when x= 1, the function f(x)=lnx-

1

n

xn+

1

nitrogen

-2 takes the maximum value,

That is g(n)= 1

1

nitrogen

-

1

n

-2,

Make t=

1

n

, then

1

nitrogen

-

1

n

-2=t2-t-2，

∫y = T2-t-2 is the opening facing upwards, and the straight line t=

1

2

A parabola with an axis of symmetry,

So when t=

1

2

When n=2, the minimum value of g(n) is-

nine

four

So the answer is:-

nine

four