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!
 A: $$\sum_{n=1}^{k}\frac{1}{(n)(n+1)(n+2)}$$
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,
$$\frac{1}{2(k+2)}-\frac{1}{2(k+1)}-\frac{3}{4}$$
A: Calling $S_n = \sum_k^n \frac{1}{k}$ we have
$$
T_n = \frac 12 S_n - \left(S_n-1+\frac{1}{n+1}\right)+\frac 12\left(S_n-1-\frac 12+\frac{1}{n+1}+\frac{1}{n+2}\right)= \frac 14+\frac 12\left(\frac{1}{n+2}-\frac{1}{n+1}\right)
$$
A: Your partial fraction is incomplete. At least you can further decompose $\frac{1}{n(n+2)}$. Anyway the correct decomposition is $\frac{1}{2}(\frac{1}{n} + \frac{1}{n+2}) - \frac{1}{n+1}$.
Now for any $3 \le n \le k$, $\frac{1}{n}$ will appear twice with coefficient $\frac{1}{2}$ and once with coefficient $-1$, therefore they all cancel. What are left? Only $1/1, 1/2, 1/(k+1), 1/(k+2)$ have nonzero coefficients. And they are $1/2, 1/2-1, 1/2-1, 1/2$ separately.
BTW, this type of problems can be usually quickly "solved" by WolframAlpha like this.
A: This is a two step partial fraction decomposition
$$S_k=\sum_{n=1}^k \frac{1}{(n)(n+1)(n+2)}=\frac{1}{2}\sum_{n=1}^k \frac{1}{(n)(n+1)}-\frac{1}{(n+1)(n+2)}\tag{1}$$
$$S_k=\frac{1}{2}\sum_{n=1}^k \left(\frac{1}{(n)}-\frac{1}{(n+1)}\right)-\left(\frac{1}{(n+1)}-\frac{1}{(n+2)}\right)=\frac{1}{2}\sum_{n=1}^k \left(\frac{1}{(n)}-\frac{2}{(n+1)}+\frac{1}{(n+2)}\right)\tag{2}$$
Then you can write (2) in terms of the Harmonic Function $H_k=\sum_{n=1}^k \frac{1}{n}$
$$S_k=\frac{1}{2}\left( H_k-2\left( H_k+\frac{1}{k+1}-1\right)+\left( H_k+\frac{1}{k+1}+\frac{1}{k+2}-\frac{3}{2}\right) \right)\tag{3}$$
Which can immediately be simplified further as all the $H_k$ terms cancel out.
