Telescoping series

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In mathematics, telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with a succeeding or preceding term. Such a technique is also known as the method of differences.

For example, the series

$\sum_{n=1}^\infty \frac{1}{n(n+1)}$

simplifies as

$\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right)\,$

$= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots\,$

$= 1 + \left(- \frac{1}{2} + \frac{1}{2}\right) + \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots = 1. \,$

[edit] A pitfall

While telescoping is a neat technique, there are pitfalls to watch out for:

$0 = \sum_{n=1}^\infty 0 = \sum_{n=1}^\infty (1-1) = 1 + \sum_{n=1}^\infty (-1 + 1) = 1\,$

is not correct because regrouping of terms is invalid unless the individual terms converge to 0; see Grandi's series. The way to avoid this error is to find the sum of the first N terms first and then take the limit as N approaches infinity:

$\sum_{n=1}^N \frac{1}{n(n+1)} = \sum_{n=1}^N \left( \frac{1}{n} - \frac{1}{n+1} \right) \,$

$= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)\,$

$= 1 + \left(- \frac{1}{2} + \frac{1}{2}\right) + \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots + \left(-\frac{1}{N} + \frac{1}{N}\right) - \frac{1}{N+1} \,$

$= 1 - \frac{1}{N+1}\to 1\ \mathrm{as}\ N\to\infty.\,$

[edit] More examples

Many trigonometric functions also admit representation as a difference, which allows telescopic cancelling between the consequent terms.

$\sum_{n=1}^N \sin\left(n\right) = \sum_{n=1}^N \frac{1}{2} \csc\left(\frac{1}{2}\right) \left(2\sin\left(\frac{1}{2}\right)\sin\left(n\right)\right)$

$=\frac{1}{2} \csc\left(\frac{1}{2}\right) \sum_{n=1}^N \left(\cos\left(\frac{2n-1}{2}\right)-\cos\left(\frac{2n+1}{2}\right)\right)$

$=\frac{1}{2} \csc\left(\frac{1}{2}\right) \left(\cos\left(\frac{1}{2}\right)-\cos\left(\frac{2N+1}{2}\right)\right).$

Some sums of the form

$\sum_{n=1}^N {f(n) \over g(n)},$

where f and g are polynomial functions whose quotient may be broken up into partial fractions, will fail to admit summation by this method. In particular, we have

$\sum^\infty_{n=0}\frac{2n+3}{(n+1)(n+2)}$

$=\sum^\infty_{n=0}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)$

$=\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots$