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 $\infty$
 A: The main term goes to zero so fast that any crude comparison gives absolute convergence.
Moreover, the value of the series is simple to compute:
$$ \sum_{n\geq 0}\frac{n^2+1}{n!}=\sum_{n\geq 0}\frac{1}{n!}+\sum_{n\geq 0}\frac{n(n-1)+n}{n!}=\sum_{n\geq 0}\frac{1}{n!}+\sum_{n\geq 1}\frac{1}{(n-1)!}+\sum_{n\geq 2}\frac{1}{(n-2)!} = \color{red}{3e}.$$
A: 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 $3e$.
A: You made a mistake at the final step to make a conclusion. :)
Since $\lim_{n\rightarrow  \infty}\left |\frac{a_{n+1}}{a_{n}}  \right |=0$ then the series converges absolutely. 
Also, we don't know the value of the sum so it is still wrong to say it converges absolutely to $0$! The ratio test just tells us that the limit
$$\lim_{m \to \infty} \sum_{n=0}^{m}\frac{n^{2}+1}{n!}$$
exists but it says nothing about the value of the limit! :)
A: HINT Observe that 
$$ \lim_{n\to\infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}=\lim_{n\to\infty}\frac{\frac{1}{n}+\frac{2}{n^2}+\frac{2}{n^3} }{ 1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}}=\frac{0}{1}=0 $$
A: The limit of a rational function (quotient of two polynomials) at $\infty$ is the limit of the ratio  of the highest degree terms, namely
$$\lim_{n\to\infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}=\lim_{n\to\infty}\frac{n^{2}}{n^{3}}=\lim_{n\to\infty}\frac{1}{n}=0.$$
