<s>[INST] 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!
 [/INST] 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.
</s>