<s>[INST] 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 ?
 [/INST] 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.
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