11Basically what happens is this:-
\frac{x}{x+1}=1-\frac{1}{x+1}
\frac{x^2}{(x+1)(x^2+1)}=\frac{1}{x+1}-\frac{1}{(x+1)(x^2+1)}
So for n terms we have the sum as 1-\frac{1}{(x+1)(x^2+1)...(x^{2n-2}+1)}
As n goes to infinity, the left term vanishes, and we have the ans as 1.