Telescoping Series

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Some series look complicated, but they’re really simple. Consider this one:

    \[ 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} -\dots \]

It’s pretty clear that every term except the 1 cancels, so the series should converge to 1. (In fact, it’s not too hard to show this. Can you write an expression for the partial sum s_n?)

Telescoping series are similar to the one above. Take a look at the series

    \[ \sum_{n=1}^{\infty}\frac{1}{n^2+n}. \]

Using partial fractions we can rewrite the terms of the series as

    \[ \frac{1}{n^2+n} = \frac{1}{n} - \frac{1}{n+1}, \]

so the new series becomes

    \[ \sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+1}. \]

Let’s take a look at the first few terms of this series:

    \[ \sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+1} = \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\dots. \]

Check out what’s happening: all the terms cancel except for the first one (1). You might want to write down a couple more terms to convince yourself of this.

Any time you can use partial fractions to simplify the terms of a series, you should see if the series is telescoping. If it is, then it converges, and you can see what value it converges to by looking at which terms don’t cancel.

Let’s look at another example: by using partial fractions on the series

    \[ \sum_{i=1}^{\infty}\frac{6}{i^2+2i}, \]

we see that the sum is equal to

    \[ \sum_{i=1}^{\infty}\frac{3}{n}-\frac{3}{n+2}. \]

The first few terms of this series are

    \[ \left(\frac{3}{1}-\frac{3}{3}\right)+\left(\frac{3}{2}-\frac{3}{4}\right)+\left(\frac{3}{3}-\frac{3}{5}\right)+\left(\frac{3}{4}-\frac{3}{6}\right)+\dots, \]

and we can see that all terms cancel except for \frac{3}{1}+\frac{3}{2}. (You should take a minute to convince yourself that all the other terms really do cance.) Therefore this series converges to this value.