(1) proves that f (x) = ∑ (x n/n 2) converges to (-1, 1).

This can be obtained by calculating the convergence radius r = lim n 2/(n+ 1) 2 = 1.

(2) According to Taylor series, there are:

ln( 1-x)=∑(-(x)^n)/n)(x∑[- 1， 1))

(3)f '(x)=∑(x^(n- 1)/n)=-ln( 1-x)/x(x∑(- 1， 1))

Yes: f' (1-x) =-ln (x)/(1-x)

(4) Let g (x) = the left side of the equation to be proved.

g '(x)= f '(x)-f '( 1-x)+ln( 1-x)/x-ln(x)/( 1-x)

=0

Therefore, G(x) is a constant c. (x∈[0, 1))

(5) prove lim g (x->; 1-)=C=π^2/6

Only need to prove: lim ln (1-x) lnx = 0 (x->1-).

This can be proved many times by Robida's law.

The subscripts of all ∑ in the above formula are n= 1, +∞.

Inference: The value of f( 1/2) can be obtained in turn.