# 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.

• The most general formula goes like $$\sum_{k=1}^n \frac1{(k)_{j+1}}=\frac1{j}\left(\frac1{j!}-\frac1{(n+1)_j}\right)$$ where $(a)_n=\prod\limits_{j=0}^{n-1} (a+j)$ is the Pochhammer symbol. Dec 26, 2011 at 14:26
• @J.M. I have derived this general result in a little more generality. And thanks for bringing this cute notation to us (it is new atleast for me!)
– user21436
Dec 26, 2011 at 15:12
• @J.M., thanks for presentation of the formula and notation of Pochhammer symbol Dec 26, 2011 at 15:22
• @J.M.: Why not make this an answer? It's perfectly valid and interesting. Dec 26, 2011 at 20:29
• @Adrian: It seems to me that Didier and N.S. have done it so I don't have to. ;) Dec 26, 2011 at 23:55

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, $$k\,S_{k}^N=\frac1{1\cdot2\cdots k}-\frac1{(N+1)\cdot(N+2)\cdots(N+k)}.$$

• $$\frac1{j!}\sum_{m=0}^j \binom{j}{m}\frac{(-1)^m}{k+m}=\frac1{(k)_{j+1}}$$ Dec 26, 2011 at 14:43
• @J.M. Indeed. 
– Did
Dec 26, 2011 at 14:49
• @DidierPiau: The very best way to do this is to use the beta function and switch the order. The beta function evaluated at integers is quite literally an integral encoding the information of the partial sum decomposition. You could also arrive at the beta function by considering the generating series, taking the derivative and noticing it is $x^n(1-x)^m$. (And then integrating from $0$ to $1$) Dec 26, 2011 at 17:48
• @EricNaslund: The very best way... Who says so? Obviously, different levels of mathematical sophistication are possible here, and several of them might have some merits.
– Did
Dec 26, 2011 at 18:14
• @Eric: May I encourage you to post the beta function route as a separate answer? :) Dec 27, 2011 at 0:53

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).

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.

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

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

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)

$\displaystyle\sum_{x=1}^{n}\prod_{i=0}^{k}\frac{1}{x+i}=\frac{-k \Gamma (k+2) \Gamma (n+1)-n \Gamma (k+2) \Gamma (n+1)-\Gamma (k+2) \Gamma (n+1)+k \Gamma (k+n+2)+\Gamma (k+n+2)}{k \Gamma (k+2) \Gamma (k+n+2)}$

$\Gamma(n+1)=n!$

• I think the OP wants to generalize the identity he has posted to the case where the denominator is the product of 3 consecutive terms and therefore your answer really doesn't answer OP's problem. Also, hints such as this are best written as comments.
– user21436
Dec 26, 2011 at 14:28
• @KannappanSampath. I was editting. But, partial fractions can do this. And the A=B tecnique should be sufficient. Dec 26, 2011 at 14:35
• GarouDan, excuse me, your comment is a bit offtopic. But thanks for your attention! Dec 26, 2011 at 14:43
• @SergeyFilkin, if you need a deduction, I think the other answer is a better ones. But, maybe you want take a look in these two link in my answer. Dec 26, 2011 at 14:48
• @GarouDan, the links and Mathematica calculations are helpful, but I was need just one equality $\frac{1}{i(i+1)(i+2)} = \frac{1}{2}\left(\frac{1}{i(i+1)}-\frac{1}{(i+1)(i+2)}\right)$ Dec 26, 2011 at 15:17