Generalize the equality $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n\cdot(n+1)}=\frac{n}{n+1}$ 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.

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
$$\sum_{i=1}^n\frac{1}{i(i+1)...(i+k)}=\frac{1}{k} \left( \frac{1}{k!} - \frac{n!}{ (n+k)!} \right)$$
