Prove that $\sum\frac{n+1}{(n+2)n!}$ converges 
Show that $\displaystyle\sum\limits_{n=1}^{\infty}\frac{n+1}{(n+2)n!}$ converges, using the integral test.

I noticed that $\displaystyle\sum\frac{n+1}{(n+2)n!} = \sum\frac{(n+1)^2}{(n+2)!}$, but how do I go about integrating a factorial?
 A: If integral test is not a must, I'll always try the ratio test when there is a factorial. Let the sequence $(a_n)$ be defined by
$$a_n = \frac{(n+1)^2}{(n+2)!} \text{ for } n \geq 1$$
Then
$$\frac{a_{n+1}}{a_n} = \frac{(n+2)^2}{(n+3)!} \cdot \frac{(n+2)!}{(n+1)^2} = \left(\frac{n+2}{n+1}\right)^2 \cdot \frac{1}{n+3} \rightarrow 1 \cdot 0 = 0$$
Therefore the series converges.
A: $$
\sum_{n\ge 1}\frac{n+1}{(n+2)n!}\le \sum_{n\ge 1}\frac{2n}{n \cdot n(n-1)} \le \sum_{n\ge 1}\frac{2n}{\frac{1}{2}n^3}=4 \sum_{n\ge 1}\frac{1}{n^2}<\pi^2.
$$
A: $$ \sum_{n=1}^{\infty} \frac{n+1}{(n+2)n!}  \leq  \sum_{n=1}^{\infty} \frac{1}{n!}  = \sum_{n=1}^{\infty} \frac{x^n |_{x=1}}{n!} = \sum_{n=0}^{\infty} \frac{x^n |_{x=1}}{n!} -1 = e^x|_{x=1}-1 =e-1$$
A: It isn't the integral test, but maybe it's interesting to note that from $$\sum_{n\geq0}\frac{x^{n}}{n!}=e^{x}
 $$ we have $$\sum_{n\geq1}\frac{x^{n}}{n!}=e^{x}-1
 $$ then $$\sum_{n\geq1}\frac{x^{n+1}}{n!}=xe^{x}-x
 $$ then if we differentiate $$\sum_{n\geq1}\frac{\left(n+1\right)x^{n}}{n!}=e^{x}+xe^{x}-1
 $$ thus $$\sum_{n\geq1}\frac{\left(n+1\right)x^{n+1}}{n!}=xe^{x}+x^{2}e^{x}-x
 $$ and now if we integrate from $0
 $ to $1
 $ we get $$\sum_{n\geq1}\frac{\left(n+1\right)}{n!}\int_{0}^{1}x^{n+1}dx=\sum_{n\geq1}\frac{\left(n+1\right)}{n!\left(n+2\right)}=\int_{0}^{1}\left(xe^{x}+x^{2}e^{x}-x\right)dx=e-\frac{3}{2}.
 $$
