Find the Maclaurin series of f(x)=(arctan(x)-x)/x^3 What I think I need to to do is find a general series expansion of the function and then derive term by term to get the Macaurin series...but I'm not quite sure how to expand this function. Any help would be appreciated!
 A: $$\arctan(x)=\sum_{n\geq 0}\frac{(-1)^n}{2n+1} x^{2n+1}\tag{1} $$
$$\arctan(x)-x=\sum_{n\geq 1}\frac{(-1)^n}{2n+1} x^{2n+1}\tag{2} $$
$$\frac{\arctan(x)-x}{x^3}=\sum_{n\geq 1}\frac{(-1)^n}{2n+1} x^{2n-2}=\color{red}{\sum_{n\geq 0}\frac{(-1)^{n+1}}{2n+3}x^{2n}}\tag{3} $$
The radius of convergence (i.e. $1$) is left unchanged by our manipulations.
A: HINT:
Note that we have
$$\begin{align}
\frac{d}{dx}\arctan(x)&=\frac{1}{1+x^2}\\\\
&=\sum_{n=0}^\infty (-1)^n x^{2n} \tag 1
\end{align}$$
for $|x|<1$.  
Then, integrate term by term to arrive at a series for $\arctan(x)$.
SPOILER ALEERT:  Scroll over the highlighted area to reveal the solution

Integrating $(1)$ term-by-term reveals $$\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$$from which we have $$\begin{align}f(x)&=\frac{\arctan(x)-x}{x^3} \\\\ &=\frac{\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} -x}{x^3}\\\\&=\frac{\sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}}{x^3}\\\\&=\sum_{n=1}^\infty \frac{(-1)^n x^{2(n-1)}}{2n+1}\\\\&=\sum_{n=0}^\infty \frac{(-1)^{n+1} x^{2n}}{2n+3}\end{align}$$

