Analyze if this series converges: $\sum_{n=0}^{\infty}\frac{n^{2}+1}{n!}$
I have used ratio test: $\lim_{n\rightarrow \infty}\left |\frac{a_{n+1}}{a_{n}} \right |< 1$
$\Rightarrow$
$\lim_{n\rightarrow \infty}\left | \frac{(n+1)^{2}+1}{(n+1)!}:\frac{n^{2}+1}{n!} \right | = \lim_{n\rightarrow \infty}\frac{((n+1)^{2}+1)n!}{(n+1)!*(n^{2}+1)}$
$= \lim_{n\rightarrow \infty}\frac{(n+1)^{2}+1}{(n+1)*(n^{2}+1)} = \lim_{n\rightarrow \infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}$
The denominator is bigger than the enumerator, so we got $\infty > 1$ and thus the sequence will diverge. (Can I just say that or this requires an additional proof? We got a $n^{2}$ in the enumerator and a $n^{3}$ in the denominator...?)
Did I do it correctly?
Edit: Converges absolutely to $0$ and NOT $\infty$