What am I doing wrong in computing this Taylor Series of $f(x) = \frac{1}{4+x^{2}}$ about a = 0? $$f(x) = \frac{1}{4+x^{2}}$$
What I did was notice that this is similar to the geometric series:
$$f(x) = \frac{1}{1-x} = 1 + x^{2} + x^{3} + x^{4} + ....$$
So I altered my original function from:
$$f(x) = \frac{1}{4 + x^{2}}$$
to:
$$f(x) = \frac{1}{4(1 - (-\frac{1}{4}x^{2}))}$$
And so it would now look like:
$$f(x) = \frac{1}{4(1 - \left(-\frac{1}{4}x^{2}\right))} = 1 + \left(\frac{-1}{4}x^{2}\right) + \left(\frac{-1}{4}x^{2}\right)^{2} + \left(\frac{-1}{4}x^{2}\right)^{3} + \left(\frac{-1}{4}x^{2}\right)^{4} + .... $$ 
now to account for the 4 out front:
$$ = 4\bigg(1 + \left(\frac{-1}{4}x^{2}\right) + \left(\frac{-1}{4}x^{2}\right)^{2} + \left(\frac{-1}{4}x^{2}\right))^{3} + \left(\frac{-1}{4}x^{2}\right))^{4} + ....\bigg) $$ 
Why is this wrong?
The correct answer is suppose to be:
$$= \frac{1}{4}-\frac{1}{16}x^2+\frac{1}{64}x^4-\frac{1}{256}x^6+\frac{1}{1024}x^8+\ldots $$
Thank you

Everything is perfect, just one thing:
You have 
$$ \frac{1}{4} \cdot \frac{1}{1-\left(-\frac{1}{4} x^2\right)} $$ 
so you need to take the series and multiply by $\frac14$, not $4$
