Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum  $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^k \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}\right)$$
Then calculating a few numbers for $n$ we get:
$$\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} \right) + \left(\frac{1}{4} - \frac{1}{3} + \frac{1}{8} \right) + \left(\frac{1}{6} - \frac{1}{4} + \frac{1}{10}\right) . . . \left(\frac{1}{2n} - \frac{1}{n+1} + \frac{1}{n+2}\right)$$
The first two fractions cancel out in the first bracket and we're left with $\frac{1}{6}$, as for the second bracket the first fraction is cancelled out by the second fraction in the third bracket.
I have noticed that the first fractrion so $\frac{1}{2n}$ cancel out by every even term in the denominator for $-\frac{1}{n+1}$ so the equation becomes:
$$\left(-\frac{1}{2n+1}+\frac{1}{n+2}\right) = \left(\frac{n-1}{(2n+1)(n+2)} \right)$$
Have I approached this correctly? I would greatly appreciate some assistance on any improvements!

$$\sum_{n=1}^{k}\frac{1}{(n)(n+1)(n+2)}$$
By partial fraction decomposition,
$$\frac{1}{(n)(n+1)(n+2)}=\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}$$
$$\sum_{n=1}^{k}\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}$$
By splitting the terms and taking the constants outside,
$$\frac{H_{k}}{2}-(H_{k+1}-1)+\frac{H_{k+2}-\frac{3}{2}}{2}$$
Where $H_{k}$ denotes the k-th harmonic number.
Using the fact that, $$H_{k+a}=H_{k}+\sum_{u=a+1}^{k+a}\frac{1}{u}$$
$$H_{k+1}=H_{k}+\frac{1}{k+1}$$ $$H_{k+2}=H_{k}+\frac{1}{k+1}+\frac{1}{k+2}$$
Placing them in summation and after some simplification (which I leave to reader) we get as follows,
$$\frac{1}{2(k+2)}-\frac{1}{2(k+1)}-\frac{3}{4}$$
