Find the sum of this series? Show that the series $\displaystyle \sum_{n=0}^{\infty}\frac{x}{(1+x^2)^n} $ converges $\forall x\in \mathbb{R}$ and find the sum $\forall x\in \mathbb{R}$
So, I've shown that the series converges $\forall x\in \mathbb{R}$ and, through testing, suspect that the sum is equal to $\frac{x}{1+x^2}$. Heres what I did:
Let $s =\displaystyle \sum_{n=0}^{k}\frac{x}{(1+x^2)^n}$ = $\frac{x}{(1+x^2)}+ \frac{x}{(1+x^2)^2} + ... +\frac{x}{(1+x^2)^k} \Rightarrow ... \Rightarrow\frac{(x^2+1)^k}{x}s = (1+x^2)^{k-1}+(1+x^2)^{k-2} + ...+1 $.
Not quite sure what to do from here. Any tips ?

Alternatively: If $x=0$, then $S=0$. If $x\ne 0$, then:
$$S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n}}\Rightarrow$$
$$(1+x^2)S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x(1+x^2)+\sum_{n=1}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x^2(1+x^2)+\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^n}=x(1+x^2)+S \Rightarrow$$
$$S=\frac{x(1+x^2)}{x^2}=\frac1x +x.$$
