Uniform convergence of following series Prove that $\sum_{n=1}^\infty \frac{x^{2n}}{(1 + x + \dots + x^{2n})^2}$ converges uniformly when $x \geq 0$.
 A: By setting $f_n(x)=\frac{x^{2n}}{(1+x+\ldots+x^{2n})^2}$ we may easily check that $f_n(x)=f_n(x^{-1})$, so it is enough to prove the series is uniformly convergent over $[1,+\infty)$ or $(0,1]$. Moreover, for any $z\in\mathbb{R}^+$ we have $z+\frac{1}{z}\geq 2$, so:
$$ 0\leq f_{n}(x) = \frac{1}{\left(x^{-n}+x^{1-n}+\ldots+x^{n-1}+x^n\right)^2}\leq\frac{1}{(2n+1)^2} $$
and:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$
A: $\sum \dfrac {x^{2n}}{(1+x+...x^{2n})^2}$
Suppose $x< 1$
$\sum \dfrac {(1-x)^2x^{2n}}{(1-x^{2n+1})^2}\\
\sum \big(\dfrac {(1-x)x^{n}}{1-x^{2n+1}}\big)^2\\
\big(\dfrac {(1-x)x^{n}}{1-x^{2n+1}}\big)^2<x^{2n}$
$\sum x^{2n}$ converges when $x<1$ and the series converges by the comparison test.
Suppose $x>1$
$\sum \big(\dfrac {(x-1)x^n}{(x^{2n+1}-1)}\big)^2\\
\sum \big(\dfrac {1-x^{-1}}{x^n-x^{-n-1}}\big)^2\\$
$\big(\dfrac {1-x^{-1}}{x^n-x^{-n-1}}\big)^2<\frac{1}{x^{2n}}$
$\sum \frac{1}{x^{2n}}$ converges when $x>1$ and the series converges by the comparison test.
Suppose $x=1,$ 
$\sum \dfrac {1}{(2n+1)^2}$ converges.
