Question: > 
> Find Power Series representation of the function $f(x) = \dfrac x{2x^2 + 1}$?
> 
> 
> 

I'm not sure how to tackle this...I'm supposed to find interval of convergence.

Answer: 
A variation: From the *[geometric series](https://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series)*
\begin{align\*}
\frac{1}{1-x}=\sum\_{n=0}^\infty x^n\qquad\qquad\qquad |x|<1
\end{align\*}

> 
> we obtain
>  \begin{align\*}
> \frac{x}{2x^2+1}&=\frac{x}{1-\left(-2x^2\right)}\\
> &=x\sum\_{n=0}^\infty (-2x^2)^n\qquad\qquad &|-2x^2|&<1\\
> &=\sum\_{n=0}^\infty (-2)^nx^{2n+1} &|x|&<\frac{1}{\sqrt{2}}\\
> \end{align\*}
> 
> 
>
$\sum x^n = \frac {1}{1-x}$ when $x$ is in the radius of convergence

$\sum (-x)^n = \frac {1}{1+x}\\
\sum (-x^2)^n =\sum (-1)^nx^{2n} = \frac {1}{1+x^2}\\
\sum (-1)^n(\sqrt 2 x)^{2n} =\sum (-1)^n(2^n) x^{2n} \frac {1}{1+2x^2}\\
x\sum (-1)^n(2^n)x^{2n} = \frac {x}{1+2x^2}\\
\sum (-1)^n(2^n)x^{2n+1} = \frac {x}{1+2x^2}$

The series conveges by the root test if:

$\lim\_\limits{n\to \infty}|\sqrt[n]{a\_n} x|<1$

$|\sqrt[n]{a\_n}| = |2^{\frac {n}{2n+1}} x|<1$

$|x|<\frac {\sqrt 2}{2}$