# Maclaurin Series Question: $\frac{x}{\sqrt{4+x^2}}$

How do I obtain the Maclaurin series representation of:

$$\frac{x}{\sqrt{4+x^2}}$$

I know I should use a well known Maclaurin Series and work my way back to the representation of this problem, but not sure what to compare it to.

-

Note that the function is $$\frac{1}{2}x\left(1 + \left(\frac{x}{2}\right)^2\right)^{-1/2}$$
You first find the Maclaurin series for $$f(x) = (1 + (x/2)^2)^{-\frac{1}{2}}$$ and then you multiply by $x/2$. Use the formula that $$(1 + x)^{k} = \sum_{n=0}^{\infty} \binom{k}{n}x^n$$ So $$f(x) = \sum_{n=0}^{\infty} \binom{-1/2}{n}\left(\frac{x}{2}\right)^{2n}.$$ "All" that you have to do is to figure out what the binomial coefficients are. By definition $$\binom{-1/2}{n} = \frac{-\frac{1}{2}(\frac{1}{2}-1)(-\frac{1}{2}-2)\cdots(-\frac{1}{2} - n +1)}{n!}.$$ Maybe if you canculate the first, say, five terms you can see a pattern. $$\binom{-1/2}{0} = 1 \\ \binom{-1/2}{1} = -\frac{1}{2}\\ \binom{-1/2}{2} = \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}$$ Try to work out some more yourself.