Your series is a geometric series; therefore the partial sums $s_n:=\sum_{k=0}^{n-1}\ldots$ can be computed explicitly. One obtains
$s_n={x^2(1+x^2) \over 2+x^2}\ \left(1-\Bigl({-1\over 1+x^2}\Bigr)^n\right)\ ;$
so it is enough to prove that $g_n(x):={x^2\over (1+x^2)^n}$ converges to $0$ uniformly with $n\to\infty$. To this end put $x^2=:u$ and investigate
$h_n(u):=u(1+u)^{-n}\qquad (0\leq u<\infty)$
instead. It is easy to verify that for $n\geq2$ the function $h_n$ assumes its maximum at $u_n={1\over n-1}$; and the value there is
$h(u_n)={1\over n-1}\ \Bigl(1+{1\over n-1}\Bigr)^{-n}\ .$
As the second factor on the right converges to ${1\over e}$ it follows that $h_n(u)\leq {C\over n}$ for some $C$ and all $u\geq 0$, $n\geq2$.