Serlo: EN: Telescoping sums and series

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Telescoping series are certain series where summands cancel against each other. This makes evaluating them particularly easy.

Telescoping sums

Introductory example

Consider the sum

Of course, we can compute all the brackets and then try to evaluate the limit when summing them up. However, there is a faster way: Some elements are identical with opposite pre-sign.

Every two terms cancel against each other. So if we shift the brackets (associative law), we get

This trick massively simplified evaluating the series. It works for any number $n\in \mathbb{N}$ of summands:

This is called principle of telescoping sums: we make terms cancel against each other in a way that a long sum "collapses" to a short expression.

General introduction

Datei:Teleskopsumme – Definition und Erklärung.webm

Telescoping sums work like collapsing a telescope

A telescoping sum is a sum of the form $\sum _{k=1}^n(a_k-a_{k+1})$. Neighbouring terms cancel, so one obtains:

analogously,

The name "telescoping sum" stems from collapsible telescopes, which can be pushed together from a long into a particularly short form.

Partial fraction decomposition

Unfortunately, most of the sums which can be "telescope-collapsed" do not directly have the above form, but must be brought into it. The following is an example:

The does not look like a telescoping sum: there is just one fraction. but there is a trick, which makes it a telescoping sum. For each $k\in \mathbb{N}$ we have:

So

And this is a telescoping sum. Who would have guessed that ?! Vorlage:Smiley The re-formulation $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ has a name: it is called partial fraction decomposition. A fraction with a product in the denominator is split into a sum, where each summand has only one factor in the denominator. This trick can serve in a lot of cases for turning a sum over fractions into a telescoping sum.

Telescoping series

Introductory example

What happens for infinitely many summands? Consider the series

The partial sums of this series are telescoping sums: For all $n\in \mathbb{N}$, there is:

So the limit amounts to

General introduction

Telescoping series are series whose sequences of partial sums are telescoping sums. They have the form $\sum _{k=1}^{\mathrm{\infty }}(a_k-a_{k+1})$. Their partial sums have the form

To see whether a telescoping series converges, we need to investigate whether the sequence ${(a_1-a_{n+1})}_{n\in \mathbb{N}}$ converges. This sequence in turn converges, if and only if $(a_n{)}_{n\in \mathbb{N}}$ converges. If $a$ is the limit of $(a_n{)}_{n\in \mathbb{N}}$ , then the limit of the telescoping series amounts to

If $(a_n{)}_{n\in \mathbb{N}}$ diverges, then $(a_1-a_{n+1}{)}_{n\in \mathbb{N}}$ diverges, as well and the entire telescoping series diverges. Analogously, the series $\sum _{k=1}^{\mathrm{\infty }}(a_{k+1}-a_k)=(a_{n+1}-a_1{)}_{n\in \mathbb{N}}$ converges, if we can show that $(a_n{)}_{n\in \mathbb{N}}$ converges. In that case, the limit is

Series are sequences and vice versa

In the beginning of the chapter, we have used that a series is actually nothing else than a sequence (of partial sums) Conversely, any sequence $(a_n{)}_{n\in \mathbb{N}}$ can be made a series if we write it as a telescoping series: We can write

So a sequence element can be written as

The sequence $(a_n{)}_{n\in \mathbb{N}}$ can hence be interpreted as a series $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ , where the "series" is seen identical with "sequence of partial sums", here.

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