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
- <math>\sum_{n=1}^\infty \frac{1}{n(n+1)}</math>
simplifies as
- <math>
\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{(n+1)}\,</math>
- <math>= \left(1 - \frac{1}{2}\right)
+ \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots\, </math>
- <math>= 1 + \left(- \frac{1}{2} + \frac{1}{2}\right)
+ \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots = 1. \,</math>
A pitfall
While telescoping is a neat technique, there are pitfalls to watch out for:
- <math>0 = \sum_{n=1}^\infty 0 = \sum_{n=1}^\infty (1-1) = 1 + \sum_{n=1}^\infty (-1 + 1) = 1\,</math>
is not correct because regrouping of terms is invalid unless the individual terms converge to 0. 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:
- <math>
\sum_{n=1}^N \frac{1}{n(n+1)} = \sum_{n=1}^N \frac{1}{n} - \frac{1}{(n+1)}\,</math>
- <math>= \left(1 - \frac{1}{2}\right)
+ \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)\, </math>
- <math>= 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} \,</math>
- <math>= 1 - \frac{1}{N+1}\to 1\ \mathrm{as}\ N\to\infty.\,</math>
More examples
- Many trigonometric functions also admit representation as a difference, which allows telescopic cancelling between the consequent terms.
- <math>\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)</math>
- <math>=\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)</math>
- <math>=\frac{1}{2} \csc\left(\frac{1}{2}\right) \left(\cos\left(\frac{1}{2}\right)-\cos\left(\frac{2N+1}{2}\right)\right).</math>
- Some sums of the form
- <math>\sum_{n=1}^N {f(n) \over g(n)},</math>
- 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
- <math>\sum^\infty_{n=0}\frac{2n+3}{(n+1)(n+2)}</math>
- <math>=\sum^\infty_{n=0}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)</math>
- <math>=\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</math>
- <math>=\infty.</math>
- The problem is that the terms do not cancel.de:Teleskopsumme