# Find all values of x for which the infinite series $S = \sum_{n=0}^\infty\left(\frac{x^2}{x^2+1}\right)^n$ converges

Find all values of x for which the infinite series $S = \sum_{n=0}^\infty\left(\frac{x^2}{x^2+1}\right)^n$ converges, and express $S$ as a function of $x$.

I think the interval of convergence is $[-1 \quad1 ]$ , but I'm not sure how to show that. I tried using ratio and root tests but I got inconclusive results. I think maybe I have to use the geometric series but when I tried I got non real roots. I'm not sure what to do and have no answer key so any help would be appreciated.

Edit: Made a careless mistake in the root test and ended up taking the limit as $x \to \infty$ instead of $n \to \infty$

• Can you write the root test ? I think it may actually lead to a conclusion.
– Dark
Commented Apr 21, 2016 at 17:20

since $$\forall x\in\mathbb{R}, 0 \leq \frac{x^2}{x^2+1} < 1$$ since for any $\rho\in(-1,1)$ $$\sum_{n\geq 0}\rho^n = \frac{1}{1-\rho}$$ this series converges for all $x\in\mathbb{R}$
Using the $\;n\,-$ th root test:
$$\sqrt[n]{\left(\frac{x^2}{x^2+1}\right)^n}=\frac{x^2}{1+x^2}\;,\;\;\text{and since}\;\frac{x^2}{1+x^2}=q<1\;\;\forall\,x\in\Bbb R$$
$$\sum_{n=0}^\infty\left(\frac{x^2}{1+x^2}\right)^n=\frac1{1-\frac{x^2}{1+x^2}}=1+x^2$$