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 ?

$(1-a)(1+a+...+a^k)=1-a^{k+1}.$ Examples: 
$(1-a)(1+a)=(1+a)-(a+a^2)=1-a^2.$ 
$(1-a)(1+a+a^2)=(1+a+a^2)-(a+a^2+a^3)=1-a^3.$ 
$(1-a)(1+a+a^2+a^3)=(1+a+a^2+a^3)-(a+a^2+a^3+a^4)=1-a^4.$ 
So if $a\ne 1$ then $(1+a+...+a^k)=(1-a^{k+1})(1-a)^{-1}.$ 
When $x\ne 0$  apply this to $s(k,x)=x\sum_{n=0}^k (1+x^2)^{-n}$ with  $a=(1+x^2)^{-1}.$
