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.
$\endgroup$2 Answers
$\begingroup$$\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:
$\lim_\limits{n\to \infty}|\sqrt[n]{a_n} x|<1$
$|\sqrt[n]{a_n}| = |2^{\frac {n}{2n+1}} x|<1$
$|x|<\frac {\sqrt 2}{2}$
$\endgroup$ 3 $\begingroup$A variation: From the geometric series\begin{align*} \frac{1}{1-x}=\sum_{n=0}^\infty x^n\qquad\qquad\qquad |x|<1 \end{align*}
$\endgroup$ 2we 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*}
