I'm not sur but a primitive of 1/(1-x^2) is Argth(x) so by integration we have :

\sum_{n=0}to{infty} (x^(2n+1))/(2n+1) = Argth(x) so

x.\sum_{n=0}to{infty} (x^(2n))/(2n+1) = Argth(x)

x.\sum_{n=0}to{infty} ((x^2)^n)/(2n+1) = Argth(x). Take x= i/sqrt(3) such that x^2= -1/3 and you have :

i/sqrt(3).\sum_{n=0}to{infty} ((-1/3)^n)/(2n+1) = Argth(i/sqrt(3)) and finaly :

\sum_{n=0}to{infty} ((-1/3)^n)/(2n+1) = Argth(i/sqrt(3))/(i/sqrt(3))

I hope I didn't make mistake.