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Finding the explicit sum of a telescoping series with two factors in the denominator is quite easy: we split the fractions in the difference of two subpieces.

But what about 2+ factors? E.g., consider:

$$\sum\frac{1}{(2n+1)(2n+3)(2n+5)}$$

We could split it in three pieces by partial fractions, but difference of three pieces would be useless.

Suggestions? Thanks.

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This is very similar in nature to this question. Some of the answers there might be useful. –  robjohn Jan 19 '13 at 22:31

4 Answers 4

up vote 3 down vote accepted

First, note that the telescoping series method only works on certain fractions. In particular, in order for the fractions to cancel out, we need the numerators to be the same. The typical example of telescoping series (for partial fractions) is

$$ \frac {1}{n(n+1)} = \frac {1}{n} - \frac {1}{n+1} \Rightarrow \sum_{i=1}^n \frac {1}{i(i+1)} = \sum_{i=1}^n \frac {1}{i} - \frac {1}{i+1} = \frac {1}{1} - \frac {1}{n+1}$$

If the numerators do not cancel out completely, then telescoping series will no longer work. For example, since we have

$$ \frac {2}{n} - \frac {1}{n+1} = \frac {n+2}{n(n+1)},$$

We find that $\sum \frac {n+2}{n(n+1)}$ doesn't yield to the telescoping series method.

How do we extend this? The key idea is that the numerators must cancel out. For example, from $1-2+1=0$, we can create

$$ \frac {1}{n} - \frac {2}{n+1} + \frac {1}{n+2} = \frac {2}{n(n+1)(n+2)}$$

which then tells us that

$$\sum_{i=1}^n \frac {2}{i(i+1)(i+2)} = \sum_{i=1}^n\frac {1}{i} - \frac {2}{i+1} + \frac {1}{i+2} = \frac {1}{1} - \frac {1}{2} + \frac {1}{n+1} - \frac {1}{n+2}.$$

You can easily see how to create more telescoping series sums using this idea. For example, what can we do with $3-4+1 = 0$ and the fraction:

$$ \frac 3 n - \frac 4 {n+1} + \frac 1 {n+2} = \frac {2(n+3)}{n(n+1)(n+2)}?$$


Edit: So, it sounds to me that you do not really see how telescoping series works. Let me expand it out, so you could understand it better. Let's first take the typical example of $\sum \frac {1}{i(i+1)}$. We can see that $\frac {1}{i(i+1)} = \frac {1}{i} - \frac {1}{i+1}$. What does this tell us? We have

$$\begin{array}{llllllllll} \frac {1}{1 \times 2} & = & \frac {1}{1} & - \frac {1}{2} \\ \frac {1}{2 \times 3} & = & & +\frac {1}{2} & - \frac {1}{3} \\ \frac {1} {3 \times 4} & = & & & + \frac {1}{3} & - \frac {1}{4} \\ \vdots & = \vdots \\ \frac {1}{(n-1) \times n} = &&&&&&& +\frac {1}{n-1} &- \frac {1}{n}\\ \frac {1}{n \times (n+1)} = &&&&&&& +\frac {1}{n} &- \frac {1}{n+1}\\ \end{array}$$

Now, summing up the LHS, we get $\sum_{i=1}^n \frac {1}{i(i+1)}$ as intended. Let's sum up the RHS according to each column. Then, we clearly see that lots of things cancel out, leaving us with $\frac 1 1 - \frac 1 {n+1}$.

We know do the same with $\frac {1}{n} - \frac {2}{n+1} + \frac {1}{n+2} = \frac {2}{n(n+1)(n+2)}$. Let's write it out as:

$$\begin{array}{llllllllll} \frac {2}{1 \times 2 \times 3} & = & \frac {1}{1} & - \frac {2}{2} & + \frac {1}{3}\\ \frac {2}{2 \times 3 \times 4} & = & & +\frac {1}{2} & - \frac {2}{3} & + \frac {2}{4}\\ \frac {2} {3 \times 4 \times 5} & = & & & + \frac {1}{3} & - \frac {2}{4} & + \frac {1}{5} \\ \vdots & = \vdots \\ \frac {2}{n \times (n+1) \times (n+2)} = &&&&&&& +\frac {1}{n-1} &- \frac {2}{n} &+ \frac {1}{n+1}\\ \frac {2}{n \times (n+1) \times (n+2)} = &&&&&&&& +\frac {1}{n} &- \frac {2}{n+1} + \frac {1}{n+2}\\ \end{array}$$

Now, can you tell what we get when we sum up the RHS over the columns? You should see that we get $\frac {1}{1} - \frac {1}{2} - \frac {1}{n+1} + \frac {1}{n+2}$.

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Note that if the denominator is squarefree with integer roots, then the coefficients sum to $0$ iff the degree of the numerator is small enough for the series to converge. This makes telescoping series pretty broadly applicable. –  Erick Wong Jan 19 '13 at 22:50
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Stupid correction to a nice answer: your example of telescoping series no longer working is wrong: $\frac{n-1}{n(n+1)}=\frac{2}{n+1}-\frac{1}{n}$. ;) –  Andrea Orta Jan 19 '13 at 23:44
    
@AndreaOrta Thanks for pointing that out! I didn't cross multiply lol. –  Calvin Lin Jan 20 '13 at 16:21
    
You are saying that, having three (or n) integers that sum to zero, and 3 (or n) firts-degree polynomials, the construction of a telescoping series is guaranteed? Regarding your question, I suspect of having some lacks in hing-school algebra: indeed, some algebraic steps are still obscure. –  MadHatter Jan 20 '13 at 18:01
    
@UzzoloDivendetta I interpreted your question in a more general setting, whereas everyone just solved your telescoping series. In the first place, do you understand how a telescoping series 'works', and why it works? Can you clarify what you mean by algebraic steps - Are you referring to the adding of fractions by making a common denominator? –  Calvin Lin Jan 20 '13 at 18:05

Write $$\dfrac1{(2n+1)(2n+3)(2n+5)} =\\ \dfrac1{2n+3} \left(\frac{1}{(2n+1)(2n+5)}\right) =\\ \dfrac1{4(2n+3)} \left(\frac{1}{2n+1}-\frac{1}{2n+5}\right) =\\ \dfrac14 \left(\frac{1}{(2n+1)(2n+3)}-\frac{1}{(2n+3)(2n+5)}\right) =a_n-a_{n+1},\\ \text{for } \ a_n=\dfrac14 \left(\frac{1}{(2n+1)(2n+3)}\right) $$ to make it easy (telescopic with two factors).

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This example is certainly useful, but has i bit of "a posteriori" flavor. Actually, you already knew the necessity of pulling out (2n+3) factor. If you pull out, for example, one of the other two, you don't obtain two successive terms of a sequence. I'm right? –  MadHatter Jan 20 '13 at 17:46

$$\dfrac1{(2n+1)(2n+3)(2n+5)} = \dfrac18 \dfrac1{2n+1} - \dfrac14 \dfrac1{2n+3} + \dfrac18 \dfrac1{2n+5}$$ The above is obtained using partial fractions. The idea behind partial fractions in this case is to write $\dfrac1{(2n+1)(2n+3)(2n+5)}$ as $$\dfrac{A}{(2n+1)} + \dfrac{B}{(2n+3)} + \dfrac{C}{(2n+5)}$$ The goal is to find $A$, $B$ and $C$. Since $$\dfrac{A}{(2n+1)} + \dfrac{B}{(2n+3)} + \dfrac{C}{(2n+5)} = \dfrac1{(2n+1)(2n+3)(2n+5)}$$ we get that $$\dfrac{A(2n+3)(2n+5) + B(2n+1)(2n+5) + C(2n+1)(2n+3)}{(2n+1)(2n+3)(2n+5)} = \dfrac1{(2n+1)(2n+3)(2n+5)}$$ Hence, we need $A$, $B$ and $C$ such that $$A(2n+3)(2n+5) + B(2n+1)(2n+5) + C(2n+1)(2n+3) = 1$$ for all $n$. Since this is true for all $n$, plug in $n = -\dfrac12$ to get that $$8A = 1 \implies A = \dfrac18$$ Now, plug in $n = -\dfrac32$ to get that $$-4B = 1 \implies B = -\dfrac14$$ and finally, plug in $n = -\dfrac52$ to get that $$8C = 1 \implies C = \dfrac18$$ Hence, we get that $$\dfrac1{(2n+1)(2n+3)(2n+5)} = \dfrac18 \dfrac1{2n+1} - \dfrac14 \dfrac1{2n+3} + \dfrac18 \dfrac1{2n+5}$$

Hence, \begin{align} \sum_{n=1}^{N} \dfrac1{(2n+1)(2n+3)(2n+5)}& = \dfrac18 \cdot \dfrac1{3} - \dfrac14 \cdot \dfrac15 + \dfrac18 \cdot \dfrac17\\ & + \dfrac18 \cdot \dfrac1{5} - \dfrac14 \cdot \dfrac17 + \dfrac18 \cdot \dfrac19\\ & + \dfrac18 \cdot \dfrac1{7} - \dfrac14 \cdot \dfrac19 + \dfrac18 \cdot \dfrac1{11}\\ & + \dfrac18 \cdot \dfrac1{9} - \dfrac14 \cdot \dfrac1{11} + \dfrac18 \cdot \dfrac1{13}\\ & + \cdots \end{align} \begin{align} \sum_{n=1}^{N} \dfrac1{(2n+1)(2n+3)(2n+5)} & = \dfrac18 \cdot \dfrac13 + \left( \dfrac18 - \dfrac14\right)\cdot \dfrac15 + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac17\\ & + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac19 + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac1{11}\\ & + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac1{2n+1} + \left( \dfrac18 - \dfrac14 \right) \cdot \dfrac1{2n+3} + \dfrac18 \cdot \dfrac1{2n+5}\\ & = \dfrac18 \cdot \dfrac13 + \left( \dfrac18 - \dfrac14\right)\cdot \dfrac15 + \left( \dfrac18 - \dfrac14 \right) \cdot \dfrac1{2n+3} + \dfrac18 \cdot \dfrac1{2n+5}\\ & = \dfrac1{24} - \dfrac1{40} - \dfrac18 \cdot \dfrac1{2n+3} + \dfrac18 \cdot \dfrac1{2n+5}\\ & = \dfrac1{60} - \dfrac14 \cdot \dfrac1{(2n+3)(2n+5)} \end{align}

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@UzzoloDivendetta Partial fraction decomposition ;) –  N. S. Jan 19 '13 at 22:25
    
@UzzoloDivendetta There are different ways to get the coefficients $A$, $B$ and $C$. One brute force way would have been to expand out $$A(2n+3)(2n+5) + B(2n+1)(2n+5) + C(2n+1)(2n+3) = 1 \,\,\,\,\,\, (\star)$$ i.e. get $$4(A+B+C)n^2 + (16A + 12B + 8C)n + (15A+5B+3C) = 1$$ and match the coefficients of $n^2$, $n$ and $1$ i.e. we would get $4(A+B+C)= 0 $ and $(16A + 12B + 8C) = 0$ and $(15A+5B+3C) = 1$ and solve this linear system to get $A$,$B$ and $C$. The other way is to as follows: If you have the partial praction decomposition as $$\sum_{k=1}^{N} \dfrac{a_k}{(n+b_k)}$$ multiply out the –  user17762 Jan 20 '13 at 18:37
    
denominators to get a polynomial of the form $$\sum_{k=1}^N \left(a_k \prod_{j=1, j \neq k}^N (n+b_j) \right)$$ and note that the coefficient of $a_i$ will be non-zero when we plug in $n = -b_i$, while the coefficients of the rest of the $a_i$'s will be zero, when we plug in $n=-b_i$. –  user17762 Jan 20 '13 at 18:39
    
Ok, I got the technique. The hard part is regrouping, as a matter of fact... –  MadHatter Jan 20 '13 at 19:07
    
@UzzoloDivendetta But you don't need to regroup to determine $A$, $B$ and $C$, if you plug in $n=-b_j$ that is the idea. –  user17762 Jan 20 '13 at 19:08

$$\dfrac1{(2n+1)(2n+3)(2n+5)} = \dfrac18 \left( \dfrac1{2n+1} - \dfrac1{2n+3} \right) -\dfrac{1}{8}\left( \dfrac1{2n+3} - \dfrac1{2n+5} \right)$$

Each bracket is telescopic.

The reason why it (always) turns into a telescopic sum, is simple to understand: check the answer by Pambos. It is actually the same thing what you get if you decompose each of his fractions into partial fractions. But as soon as you realize that this is the case, all you need is to do the complete PFD and group them the right way...

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