Is $2|x|/(1+x^{2})<1$ true if $|x|<1$? [Solved] I was asked to show that the series 
$$\sum_{k\geq 0}a_{k}\left ( \frac{2x}{1+x^{2}} \right )^{k}$$
is convergent for $x\in (-1,1)$, where $\{ a_{k}\}_{k\geq 0}$ is a real bounded sequence. 
My main question is, how do I know that $|2x/(1+x^{2})|<1$, that is, $2|x|/(1+x^{2})<1$ is true for $|x|<1$? 
EDIT: Thanks for your answers. It took time to find it out on my own. I hope this too is correct. Let $y(x)=2x/(1+x^{2})$. Since $y'(x)>0$ for all $x\in (-1,1)$ then $y(x)$ is strictly increasing for all $x\in (-1,1)$. Hence $y(-1)<y(x)<y(1)$ for all $x\in (-1,1)$. Since $y(-1)=-1$ and $y(1)=1$, then $| y(x)|<1$ is true for $|x|<1$.

The claim is true if $2|x| < 1+x^2$ which is true by AM-GM inequality, because equality occurs only when $x^2=1$. Because $|x|<1 \implies x\neq 1$ the inequality is strict. The AM-GM inequality holds for the domain of positive reals, but all terms involved are positive even when $x$ is negative so the above inequality is true for all reals except for $x=\pm{1}$.
