A problem of UMCYM

$\sum\limits_{n=1}^{2017} (-1)^{n+1} \frac{n^2+n+1}{n!}$

$=\sum\limits_{n=1}^{2017} (-1)^{n+1}(\frac{n^2}{n!}+\frac{n+1}{n!})$

$=\sum\limits_{n=1}^{2017} (-1)^{n+1}(\frac{n}{(n-1)!}+\frac{n+1}{n!})$

$=(\frac{1}{0!}+\frac{2}{1!})-(\frac{2}{1!}+\frac{3}{2!})+(\frac{3}{2!}+\frac{4}{3!})-(\frac{4}{3!}+\frac{5}{4!})+...-(\frac{2016}{2015!}+\frac{2017}{2016!})+(\frac{2017}{2016!}+\frac{2018}{2017!})$

$=\boxed{1+\frac{2018}{2017!}}$