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 ?
 A: Let $x$ be a non-zero real number.
Then given sum is $x\cdot(1+\frac 1{1+x^2}+\frac 1{{(1+x^2)}^2}+\frac 1{{(1+x^2)}^3}+\cdot\cdot\cdot).$
Observe that $1+\frac 1{1+x^2}+\frac 1{{(1+x^2)}^2}+\frac 1{{(1+x^2)}^3}+\cdot\cdot\cdot$ is a geometric series with first term $a=1$ and common ratio $r=\frac 1{1+x^2} \implies |r|=\frac 1{1+x^2} \lt 1$.
Thus this geometric series converges to limit $\frac a{1-r}=\frac 1{1-\frac 1{1+x^2}}=\frac {1+x^2}{x^2}.$
Thus given sum is $x\cdot(1+\frac 1{1+x^2}+\frac 1{{(1+x^2)}^2}+\frac 1{{(1+x^2)}^3}+\cdot\cdot\cdot) =x \cdot \frac {1+x^2}{x^2}=\frac 1x + x.$
In case of $x=0$, every term of the series is zero. Thus whole sum is also zero.
A: I think you can use Geometric series test. Take x outside and you can see 
term  $\sum_{n=0}^\infty \frac{1}{(1+x^2)^n}$ converges when $\frac{1}{|x^2+1 |}<1 $
So by the Geometric series test, you can find out common ratio $\frac{1}{x^2+1}$. 
which is $\le$ 1. Hence proved. 
A: $(1-a)(1+a+...+a^k)=1-a^{k+1}.$ 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  $a=(1+x^2)^{-1}.$
A: Alternatively: If $x=0$, then $S=0$. If $x\ne 0$, then:
$$S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n}}\Rightarrow$$
$$(1+x^2)S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x(1+x^2)+\sum_{n=1}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x^2(1+x^2)+\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^n}=x(1+x^2)+S \Rightarrow$$
$$S=\frac{x(1+x^2)}{x^2}=\frac1x +x.$$
