Three telescoping sums

1. $k\cdot k!=(k+1-1)k!=(k+1)!-k!$. The sum telescopes:
$$\sum_{k=1}^nk\cdot k!=(n+1)!-1.$$

2. $\dfrac k{(k+1)!}=\dfrac{(k+1)-1}{(k+1)!}=\dfrac1{k!}-\dfrac1{(k+1)!}$, so
$$\sum_{k=1}^n\frac k{(k+1)!}=1-\frac1{(n+1)!}.$$

3. $\dfrac1{k(k+1)(k+2)}=\dfrac12\left(\dfrac1{k(k+1)}-\dfrac1{(k+1)(k+2)}\right)$, so
$$\sum_{k=1}^n\frac1{k(k+1)(k+2)}=\frac12\left(\frac12-\frac1{(n+1)(n+2)}\right)=\frac{n(n+3)}{4(n+1)(n+2)}.$$

Check. $n=1$: (1) $1=2!-1$; (2) $\frac12=1-\frac12$; (3) $\frac16=\frac{4}{4\cdot2\cdot3}$ ✓.

Lesson. Make terms telescope by adding and subtracting a suitable quantity.