$f(x) = \frac{x}{2x^2+1}$ Power Series Interval of Convergence Alright, so I have a homework problem and I'm struggling on finding the interval of convergence/just need some reinforcement on this concept to garner a stronger grasp on how to solve these problems. So here is the problem: $f(x) = \frac{x}{2x^2+1}$ 
So to start I rewrote: $f(x) = x\frac{1}{1-(-2x^2)}$ and then translated into series form:
$x\sum{(-1)^n(-2x^2)^n}$
Which then becomes: $\sum{(-1^n)(-2x^{2n+1})}$ Although <-- Please make sure I did that correctly, I was a lil "iffy" on that.
Now, I need to find the interval of convergence, which basically means that the x value must always be less than 1 for it to converge, but I'm curious as to how I would do that, would it be?
$$-2x^{2n+1}<1$$
The problem here is the "n" value, I don't know how to deal with it or if I'm attempting to find the interval of convergence wrong or like if it should be:
$$-2x^{2}<1$$
And then just solve the inequality. So yeah, just need help making sure that I can find that interval if you see any other problems with my work/thought process above please tell me! Thanks!
 A: The problem is to determine the power series and interval of convergence for
$$f(x) = \frac{x}{2x^2+1} \tag{1}\label{eq1}$$
As you stated, $f(x) = x\frac{1}{1-(-2x^2)}$. Since the $x$ factor doesn't affect the summation of the fraction expansion, it's better to not consider it when determining the radius of convergence. Instead, note that $\frac{1}{1-(-2x^2)}$ is in the form of the sum of an infinite series, i.e., $\frac{1}{1 - r}$, where $a = 1$ and $r = -2x^2$. This will be true for all cases where $\left| r \right| \lt 1$, i.e., $\left| -2x^2 \right| \lt 1 \; \Rightarrow \; \left| x \right| \lt \frac{1}{\sqrt{2}}$. You had the right idea with your $-2x^2 \lt 1$ statement, except you are missing using the absolute value. To confirm that you can't use any larger $x$, note that in the infinite sum $\sum_{n = 0}^{\infty}(-2x^{2})^n$, if $2x^2 \ge 1$, then the magnitude of each term is $\ge 1$, so the sum cannot converge since for any sequence $b_i$, note that $\sum_{n=0}^{\infty}b_n$ can only possibly converge if $\lim_{n \to \infty}b_n = 0$. In addition, this radius of convergence is given, as I've stated it, in Wikipedia's Geometric series section about the Taylor series.
Overall, you end up with
$$f(x) = \sum_{n = 0}^{\infty}(-2)^n x^{2n + 1} \tag{2}\label{eq2}$$
for all $\left| x \right| \lt \frac{1}{\sqrt{2}}$.
