TITLE: Generalize the equality $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n\cdot(n+1)}=\frac{n}{n+1}$
QUESTION [5 upvotes]: I'm reading a book The Art and Craft of Problem Solving. I've tried to conjecture a more general formula for sums where denominators have products of three terms. I've "got my hands dirty", but don't see any regularity in numerators.
Please, write down your ideas.

REPLY [2 votes]: Too long for a comment, some of the answers posted are longer than they need. You don't need the full partial fraction decomposition or mathematica/calculators:
$$\frac{1}{i(i+k)}= \frac{1}{i} \left( \frac{1}{i} - \frac{1}{i+k} \right) \,.$$
Thus
$$\frac{1}{i(i+1)...(i+k)}= \frac{1}{k} \left( \frac{1}{i(i+1)...(i+k-1)} - \frac{1}{(i+1)(n+2)....(i+k)} \right) \,.$$
Summing, the right side is telescopic, thus
$$\sum_{i=1}^n\frac{1}{i(i+1)...(i+k)}=\frac{1}{k} \left( \frac{1}{k!} - \frac{n!}{ (n+k)!} \right)$$