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.
 A: Assuming the $n$th term of the series $S_k^N$ you want to sum is
$$
x_k^n=\frac1{n(n+1)\cdots(n+k)},
$$
the standard approach is to decompose it as a sum
$$
x_k^n=\sum\limits_{i=0}^k\frac{c_{i,k}}{n+i},
$$
for some coefficients $(c_{k,i})_i$ and to proceed from there. For example,
$$
x_2^n=\frac1{n(n+1)(n+2)}=\frac12\cdot\frac1n-\frac1{n+1}+\frac12\cdot\frac1{n+2},
$$
hence
$$
S_2^N=\sum_{n=1}^Nx_2^n=\frac12\sum_{n=1}^N\frac1n-\sum_{n=2}^{N+1}\frac1{n}+\frac12\sum_{n=3}^{N+2}\frac1{n},
$$
and from there one can deduce a closed form expression of $S_2^N$.
Another method is to note that
$$
(k+1)x_{k+1}^n=x_k^n-x_ k^{n+1},
$$
hence
$$
(k+1)S_{k+1}^N=x_k^1-x_k^{N+1},
$$
that is,
$$
k\,S_{k}^N=\frac1{1\cdot2\cdots k}-\frac1{(N+1)\cdot(N+2)\cdots(N+k)}.
$$
A: If I understand your statement about three terms in the denominator correctly it seems that you are looking for a closed formula for $$\sum_{i=1}^{n-2} \frac{1}{i(i+1)(i+2)}$$
Note that $$\frac{1}{i(i+1)(i+2)} = \frac{1}{2}\left(\frac{1}{i(i+1)}-\frac{1}{(i+1)(i+2)}\right)$$
Combining this with the result you already got you can easily derive a formula for the sum.
If you want to see more results of that kind I'd recommend once more Konrad Knopp's book about infinite sequences and series which has many examples of explicitly calculated limits obtained by all kinds of manipulations.
(It should also be quite obvious how to generalize this further to sums with more factors in the denominator).
A: I have the following recipe in mind, see how far it helps (leave pointers in this regards as comments):
Let $a_1, a_2, a_3, \cdots, a_n, \cdots$ be the terms of an $A.P$. Let $d$ be the common difference of the given $A.P$. We are interested to find the sum for some $r \in \mathbb{N}$. $$\sum_{k=1}^n \dfrac{1}{a_k a_{k+1} \cdots a_{k+r-1}}$$
Let us denote the sum of the series as $S_n$ and the n-th term of the series, $T_n$. Consider the following definition of the new entity that I'll call $V_n$. (This is not at all a mystery: $V_n$ is obtained by dropping the first of the $r$ entries in the denominator of $T_n$)
$$V_n:=\dfrac{1}{a_{n+1} \cdots a_{n+r-2} a_{n+r-1}}$$
Therefore, $$V_{n-1}:=\dfrac{1}{a_n  \cdots a_{n+r-3} a_{n+r-2}}$$
Now (I'll leave the computation that goes here!), you'll have $$V_n-V_{n-1}=T_n(a_n-a_{n+r-1})$$ $$T_n=\dfrac{1}{d(r-1)} \cdot (V_{n-1}-V_n)$$Substituting various values for $n$, you have equations for $T_1, T_2, \cdots, T_n$. Adding these, and noting that common difference is the difference between two consecutive terms taken in an (appropriate!) order, you have, $$S_n=\dfrac{1}{(r-1)(a_2-a_1)}(\dfrac{1}{a_1\cdots a_{r-1}}-\dfrac{1}{a_{n+1} \cdots a_{n+r-1}})$$
For the problem at hand, set $a_1=1,~d=1,~r=3 $. You'll get $$\sum_{k=1}^n{\dfrac{1}{k(k+1)(k+2)}}=\dfrac{1}{4}-\dfrac{1}{(n+1)(n+2)}$$
Have fun doing for $r=4, \cdots$. Hope this helps.
A: 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
$$\sum_{i=1}^n\frac{1}{i(i+1)...(i+k)}=\frac{1}{k} \left( \frac{1}{k!} - \frac{n!}{ (n+k)!} \right)$$
A: This kind of technique calls partial fractions and can be applied to several similar problems.
If you want a very powerfull techinque, try to read a book called A=B.
Using Mathematica (not only three terms, but using k ones)
$\displaystyle\sum_{x=1}^{n}\prod_{i=0}^{k}\frac{1}{x+i}=\frac{-k \Gamma (k+2) \Gamma (n+1)-n \Gamma (k+2) \Gamma (n+1)-\Gamma (k+2)
   \Gamma (n+1)+k \Gamma (k+n+2)+\Gamma (k+n+2)}{k \Gamma (k+2) \Gamma
   (k+n+2)}$
$\Gamma(n+1)=n!$
