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 $|x|<1$.
 A: Looking at positive values only, $\frac{2x}{1 + x^2} <1$ leads to $2x < 1 + x^2$ which leads to $0 < x^2 - 2x + 1 = (x - 1)^2$--this is true for any value of $x \neq 1$--that is all values of $0 \leq x < 1 \cup 1 < x < \infty$.  If we consider negative values then we have that $\frac{-2x}{1 + x^2} < 1$ which leads to $0 < x^2 + 2x + 1 = (x + 1)^2$.  In this case we see the inequality is true for $-\infty < x < -1 \cup -1 < x \leq 0$.
By combining these two intervals we get that:
$$
\frac{2|x|}{1 + x^2} < 1 \text{ when } x \in (-\infty, -1)\cup(-1, 1)\cup (1, +\infty)
$$
Otherwise, when $x = \pm 1$, $\frac{2|x|}{1 + x^2} = \frac{2}{2} = 1 \nless 1$.
A: 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 $x=\pm{1}$.
