Analyze if this series converges: $\sum_{n=0}^{\infty}\frac{n^{2}+1}{n!}$ 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 $\infty$**

Just to provide a concrete underpinning to this general reasoning: $$\sum_{n=0}^{\infty}\frac{1}{n!}x^n=e^x$$ Differentiate and then multiply by $x$: $$\sum_{n=0}^{\infty}\frac{n}{n!}x^n=xe^x$$ Differentiate and then multiply by $x$: $$\sum_{n=0}^{\infty}\frac{n^2}{n!}x^n=x(x+1)e^x$$ Adding the first and last equations: $$\sum_{n=0}^{\infty}\frac{n^2+1}{n!}x^n=(x^2+x+1)e^x$$ So the value of your sum is $3e$.