# Telescoping series

(Redirected from Telescoping cancellation)

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

[itex]\sum_{n=1}^\infty \frac{1}{n(n+1)}[itex]

simplifies as

[itex]

\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{(n+1)}\,[itex]

[itex]= \left(1 - \frac{1}{2}\right)

+ \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots\, [itex]

[itex]= 1 + \left(- \frac{1}{2} + \frac{1}{2}\right)

+ \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots = 1. \,[itex]

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

[itex]0 = \sum_{n=1}^\infty 0 = \sum_{n=1}^\infty (1-1) = 1 + \sum_{n=1}^\infty (-1 + 1) = 1\,[itex]

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:

[itex]

\sum_{n=1}^N \frac{1}{n(n+1)} = \sum_{n=1}^N \frac{1}{n} - \frac{1}{(n+1)}\,[itex]

[itex]= \left(1 - \frac{1}{2}\right)

+ \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)\, [itex]

[itex]= 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} \,[itex]

[itex]= 1 - \frac{1}{N+1}\to 1\ \mathrm{as}\ N\to\infty.\,[itex]de:Teleskopsumme

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