# Stumped by this Maclaurin series question

It says to find the Maclaurin series in sigma form and the interval on which it converges for

$$\dfrac{7x^4}{2+3x^2} \qquad \text{if} \qquad \frac{1}{1-x} = \sum_{k=0}^{\infty}\, x^k, \;\; -1 \lt x \lt 1.$$

I am confused as to what I'm supposed to do with the $\dfrac{7x^4}{2+3x^2}$. Am I supposed to take the derivative of that? And what is all that stuff to the right of it? I think that's to determine the interval which it converges on but I'm unsure.

Image from the original question:

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The stuff to the right of it is an indication you are supposed to use to solve the question asked. – Did May 24 '11 at 10:13

First of all, you have to find the Maclaurin series of $\dfrac 1{2+3x^2}$ (then it will be easy to find the series of the given function). We can write $\dfrac 1{2+3x^2} = \dfrac 12\dfrac 1{1+\frac{3x^2}2}$. Now, you can expand this fraction thank to the given result.

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Hint 1: replace $x$ by $-ax^{2}=(-1)ax^{2}$ in

$$\frac{1}{1-x}=\sum_{k=0}^{\infty }x^{k},\qquad \left\vert x\right\vert <1$$

to get

$$\frac{1}{1+ax^{2}}% =\sum_{k=0}^{\infty }\left( -1\right) ^{k}a^{k}x^{2k},\qquad \left\vert -ax^{2}\right\vert <1\Leftrightarrow \left\vert x\right\vert <% \frac{1}{\sqrt{a}},a>0$$

Hint 2: note that

$$\frac{7x^{4}}{2+3x^{2}}=\frac{7}{2}x^{4}\left( \frac{1}{1+\frac{3}{2}x^{2}}% \right) .$$

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