Telescoping Series

Calculus 2

Telescoping Series

A telescoping series is a series in which most of the terms cancel, leaving only some of the first and last terms. Similar to how a telescope collapses on itself, the inner terms of a telescoping series cancel each other.

For example, $\sum_{n=0}^\infty \left( \frac{1}{n+1} - \frac{1}{n+2} \right) = 1$ as:

\begin{aligned} S_n &= \left(\frac{1}{1} - \cancel{\frac{1}{2}}\right) + \left(\cancel{\frac{1}{2}} - \cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}} - \cancel{\frac{1}{4}}\right) + \cdots + \left(\cancel{\frac{1}{n-2}} - \cancel{\frac{1}{n-1}}\right) + \left(\cancel{\frac{1}{n-1}} - \frac{1}{n}\right) \\ &= \frac{1}{1} - \frac{1}{n} \\ \lim_{n \to \infty} S_n &= 1. \end{aligned}

Typically, partial fraction decomposition is required to determine the sum of a telescoping series. Note that the decomposition needs to result in a difference of terms for the telescoping series to converge.

Example 1

Evaluate $\sum_{k=0}^\infty \frac{1}{k^2+5k+6}$ .

View Answer

\frac{1}{2}

Example 2

Evaluate $\sum_{n=4}^\infty \frac{6}{n^2-2n}$ .

View Answer

\frac{5}{2}

If you are asked to compute the exact sum of a series, the series will most likely be either a convergent geometric series or a telescoping series.

Later, we will learn how to find the exact sum using power series.

Root Test

Calculus 3