Telescoping series
From Academic Kids
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.
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>
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>de:Teleskopsumme
