# Telescoping Series

Definition: A Telescoping Series is a series whose partial sums simplify to a fixed number of terms when expanded. |

Describing a telescoping series is a tad difficult, so let's look at an example, namely the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.

We know that the $n^{\mathrm{th}}$ term in the series can be obtained by the formula $\frac{1}{n(n+1)}$, and so a formula for the $n^{\mathrm{th}}$ partial sum would be $s_n = \frac{1}{1(2)} + \frac{1}{2(3)} + \frac{1}{3(4)} ... + \frac{1}{(n-1)(n)} + \frac{1}{n(n+1)}$. Of course finding the $\lim_{n \to \infty} s_n$ would be rather complicated. Instead we will use the fact that $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, and therefore we expand the partial sum $s_n$ as follows:

(1)After cancelling the middle terms we get that $s_n = 1 - \frac{1}{n+1}$. Now determining $\lim_{n \to \infty} s_n$ is much easier, in fact:

(2)Therefore the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$.

## Example 1

**Determine if the telescoping series $\sum_{n=2}^{\infty} \frac{1}{n^2 - n}$ is convergent or divergent. If the series is convergent then determine what value the series approaches.**

We first observe that $\frac{1}{n^2 - n} = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$, and therefore:

(3)Now we know that $\lim_{n \to \infty} s_n = \lim_{n \to \infty} \left ( 1 - \frac{1}{n} \right ) = 1$. Therefore this telescoping series converges to $1$.

## Example 2

**Determine if the telescoping series $\sum_{n=1}^{\infty} \left ( \ln(2n +4) - \ln(2n + 2) \right )$ is convergent or divergent. If the series is convergent then determine what value the series approaches.**

In this example, try to find a simplified formula for the $n^{\mathrm{th}}$ partial sum:

(4)Therefore when we cancel appropriate terms we get that $s_n = - \ln (4) + \ln (2n + 4)$. Now let's evaluate $\lim_{n \to \infty} \left ( -\ln (4) + \ln (2n + 4) \right )$.

(5)So $\lim_{n \to \infty} \ln (2n + 4) = \infty$, so then this series diverges to positive infinity.