For #1: It can be rewritten as $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}$
The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.
The second one $\displaystyle \sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2$
$\displaystyle e-1-(e-2)=1$
$\displaystyle \sum\limits_{j=1}^{n}{\frac{1}{j(j+1)(j+2)}}=\frac {1}{2}\sum\limits_{j=1}^{n}{\frac{(j+2)-j}{j(j+1)(j+2)}},$ hence your sum equals $\displaystyle \frac{1}{2}\sum\limits_{j=1}^{n}{\left( \frac{1}{j(j+1)}-\frac{1}{(j+1)(j+2)} \right)}=\frac{1}{2}\left( \frac{1}{2}-\frac{1}{(n+1)(n+2)} \right).$
Galactus did a nice job...but with an infinite series, whereas you wanted a finite one, yet the answer STILL is in Galactus message:
$\displaystyle \sum\limits_{k=1}^n\frac{k}{(k+1)!}=\sum\limits_{k =1}^n\frac{1}{k!}-\sum\limits_{k=1}^n\frac{1}{(k+1)!}=1-\frac{1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}$
Tonio