The question is: Determine the interval of convergence of the power series
$\sum_{n=1}^\infty\left(\frac{2n+1}{n^2+1}\right)(2x+1)^{12}$
My attempt at an answer: $u_n=\frac{(2n+1)}{(n^2+1)}(2x+1)^{12}$ Applying the ratio test: $\begin{align} \require{enclose} \frac{|u_{n+1}|}{|u_n|}&=\left|\frac{(2n+2)(2x+1)^{12}}{((n+1)^2+1)}.\frac{(n^2+1)}{(2n+1)(2x+1)^{12}}\right|\\ &=\left|\frac{(2n+2)\enclose{horizontalstrike}{(2x+1)^{12}}}{(n^2+2n+1)}.\frac{(n^2+1)}{(2n+1)\enclose{horizontalstrike}{(2x+1)^{12}}}\right|\\ &=\left|\frac{(2n+2)}{(n^2+2n+1)}.\frac{(n^2+1)}{(2n+1)}\right|\\ \end{align}$ But now I just got rid of all the $x$ components which is obviously wrong!?!.
$_{help!!!}$