Find the radius of convergence for the following infinite series:

$\displaystyle \sum_{n=1}^\infty\frac{x^{2n}}{3^n(n^2+1)}$

Answer: $\displaystyle \sqrt{3}$

Use the following formula:

$\displaystyle \left|\frac{a_{n+1}}{a_n}\right|$

My attempt:

$\displaystyle \left|\frac{x^{2(n+1)}}{3^{(n+1)}((n+1)^2+1)}\cdot \frac{3^n(n^2+1)}{x^{2n}}\right|=\left|\frac{x^{2( n+1)}}{x^{2n}}\cdot\frac{3^n(n^2+1)}{3^{(n+1)}((n+ 1)^2+1)}\right|=|x|\frac{3^n(n^2+1)}{3^{(n+1)}((n+ 1)^2+1)}$

I'm not sure if I'm on the right path. I can't seem to figure it out what to do.

I know that for convergence we need that $\displaystyle |x|<r$ and for divergence $\displaystyle |x|>r$.

Any help?