I first worked with this expression as follows: $$\frac{4}{3-2x^2} = \frac{1}{\frac{3-2x^2}{4}} = \frac{1}{1-\frac{2x^2+1}{4}}$$
Knowing that this is in the form of $\frac{1}{1-u}$ (which can be expressed as $\sum_{n=0}^\infty {u}^n$), I converted my example to: $$\sum_{n=0}^\infty {\left(\frac{2x^2+1}{4}\right)}^n; \frac{2x^2+1}{4} < 1 \rightarrow \frac{-\sqrt6}{2} < x < \frac{\sqrt6}{2}$$
Then, I worked with it as so: $$\frac{4}{3-2x^2} = \frac {4}{3} \cdot \frac{1}{1-\frac{2}{3}x^2} $$
From which I converted into the infinite series:
$$\sum_{n=0}^\infty {\frac 43\left(\frac{2x^2}{3}\right)}^n; \frac {2x^2}{3} < 1 \rightarrow \frac{-\sqrt6}{2} < x < \frac{\sqrt6}{2}$$
How can I modify one series so that can convert it to the other? I am finding it difficult to understand why these two series are the same.