Find the Maclaurin series for the function $\tan^{-1}(2x^2)$ Find the Maclaurin series for the function  $\tan^{-1}(2x^2)$
Express your answer in sigma notation, simplified as much as possible. What is the open interval of convergence of the series.
I have the correct answer, but I would like to use another method to solve this.
By taking the function and its derivative, find the sum and then taking the anti derivative. This does not yield the same answer for me.

You probably know the series for $\tan^{-1} t$. Plug in $2x^2$ for $t$.
If you do not know the series for $\arctan t$, you undoubtedly know the series for $\dfrac{1}{1-u}$. Set $u=-x^2$, and integrate term by term.
For the interval of convergence of the series for $\tan^{-1} (2x^2)$, you probably know when the series for $\tan^{-1}t$ converges. That knowledge can be readily translated to knowledge about $\tan^{-1}(2x^2)$. 
