I have to show the following series is either convergent or divergent.

$\displaystyle \sum_0^\infty \frac{(n^2 +1)^2.2^{2n-1}}{n!}$

I know this series isconvergentby the ratio test, but I am having trouble showing this proof.

So far I have:

$\displaystyle \rho = lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = lim_{n\rightarrow\infty} \frac{((n+1)^2 +1)^2.2^{2(n+1)-1}}{(n+1)!} . \frac{n!}{(n^2 +1)^2.2^{2n-1}}$

I simplified to

$\displaystyle \rho = lim_{n\rightarrow\infty} \frac{((n+1)^2 +1)^2}{(n^2+1)^2}.\frac{2^{2n+1}}{2^{2n-1}}.\frac{n!}{(n+1)n!}}$

$\displaystyle = lim_{n\rightarrow\infty} \frac{(n^2 +2n+2)^2}{(n^2+1)^2}.2^2.\frac{1}{(n+1)}$

$\displaystyle = lim_{n\rightarrow\infty} \frac{4.(n^2 +2n+2)^2}{(n^2+1)^2(n+1)}$

This is where I am stuck.

Could someone help me check if this is right so far and help me with what comes next?