# Find the sum to n terms of the series $\displaystyle \frac{1} {1.2.3.4} + \frac{1} {2.3.4.5} + \frac{1} {3.4.5.6}$

Please suggest an approach for this task.

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Type "sum 1/(n(n+1)(n+2)(n+3))" into wolfram alpha. –  Oscar Cunningham Sep 27 '10 at 13:34
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## 3 Answers

$\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{k+3}{k(k+1)(k+2)(k+3)} - \dfrac{k}{k(k+1)(k+2)(k+3)})$ $= \dfrac{1}{3}(\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$

$\sum_{k=1}^{\infty}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} \dfrac{1}{2*3} = \dfrac{1}{18}$

@moron Yes, you are right. For the first n terms:

$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$

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$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)(k+3)} = \sum_{k=1}^{n} \dfrac{1}{3} (\dfrac{1}{k(k+1)(k+2)} - \dfrac{1}{(k+1)(k+2)(k+3)})$ $= \dfrac{1}{3} [\sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$ $= \dfrac{1}{3} [\sum_{k=0}^{n-1} \dfrac{1}{(k+1)(k+2)(k+3)} - \sum_{k=1}^{n} \dfrac{1}{(k+1)(k+2)(k+3)}]$ $= \dfrac{1}{3} (\dfrac{1}{1*2*3} - \dfrac{1}{(n+1)(n+2)(n+3)})$

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+1: Even though the question says first n terms, this can easily be adapted for that. –  Aryabhata Sep 27 '10 at 14:09
@Moron Thank for your comment –  Branimir Sep 27 '10 at 14:15
@Branimir: You are welcome! And I see you joined today, welcome to this site :-) –  Aryabhata Sep 27 '10 at 14:20
so tricky!!! :) –  BBischof Sep 27 '10 at 14:21
Say n = 2, then 1/24 + 1/180 = 1/20 where as your solution is giving 1/18 - 1/72 = 1/24 ?! –  Quixotic Sep 27 '10 at 14:49
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This is similar to what has been said by Branimir, but shows how we can extend the result to $$\sum_{k=1}^n {1 \over k(k+1) \cdots (k+m)}, \qquad m \in \mathbb{N}.$$

We can build up the result from the identities $${1 \over k(k+1)} = {1 \over k} - { 1 \over k+1}, \qquad (1)$$ $${1 \over k(k+1)(k+2)} = {1 \over 2} \left( {1 \over k(k+1)} - { 1 \over (k+1)(k+2)} \right),$$ $${1 \over k(k+1)(k+2)(k+3)} = {1 \over 3} \left( {1 \over k(k+1)(k+2)} - { 1 \over (k+1)(k+2)(k+3)} \right), \quad \textrm{ etc...}$$

Write $S_1 = \sum_{k=1}^n {1 \over k(k+1)},$ $S_2 = \sum_{k=1}^n {1 \over k(k+1)(k+2)},$ etc

Summing for $S_1$ using (1) all terms on RHS cancel to get the classic $$S_1 = \sum_{k=1}^n {1 \over k(k+1)} = 1 – {1 \over n+1} = {n \over n+1}.$$ We then sum the series for $S_2$ using this result obtained for $S_1,$ and so on.

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This can be expressed by saying that "negative falling powers behave nicely with respect to the difference operator". See p.53 in "Concrete Mathematics" by Graham, Knuth & Patashnik. –  Hans Lundmark Sep 27 '10 at 15:22
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HINT $\rm\displaystyle\ \frac{1}{(k+1)(k+2)(k+3)(k+4)} = \frac{1}{6(k+1)} - \frac{1}{2(k+2)}+\frac{1}{2(k+3)}-\frac{1}{6(k+4)}$

$\rm\ f(k+1)-f(k)\: =$ above $\rm\displaystyle\ \Rightarrow\ f(k) \:=\: c_0 + \frac{c_1}{k+1}\ \:+\:\ \frac{c_2}{k+2}\ \:+\:\ \frac{c_3}{k+3}$

Calculating yields $\rm\ c_0,c_1,c_2,c_3 \ =\ 1/18,\ -1/6,\ 1/3,\ -1/6$.

For remarks on the group theory behind rational indefinite summation see my post here

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