Upper bound for $\frac{x^{2k+2}}{1-x^2} $ My book says that if $x \in (-0.5, 0.5)$, then
$$\frac{x^{2k+2}}{1-x^2} \leq \frac{1}{3\cdot 4^k}$$
I don't understand how to get such upper bound?
Take $k=0 \Rightarrow \frac{x^{2}}{1-x^2} \leq \frac{0.25}{0.75} = \frac{1}{3}$
But how to figure out the general form?

Note that if $|x| < .5 = \frac 1 2$, then
$$\frac{x^{2k + 2}}{1 - x^2} = \frac{x^2}{1 - x^2} \cdot (x^2)^k \le \frac 1 3 (x^2)^k \le \frac 1 3 \cdot \left(\frac 1 4\right)^k$$
using your previous estimate.
