Telescoping sums and series – "Math for Non-Geeks"

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

Telescoping sums

Introductory example

Consider the sum

$\begin{array}{cc}\sum _{k=1}^4(\frac{1}{k}-\frac{1}{k+1})& =(\frac{1}{1}-\frac{1}{1+1})+(\frac{1}{2}-\frac{1}{2+1})+(\frac{1}{3}-\frac{1}{3+1})+(\frac{1}{4}-\frac{1}{4+1})\\ & =(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})\end{array}$

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.

$\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)$

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

$\begin{array}{cc}& 1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+(-\frac{1}{4}+\frac{1}{4})-\frac{1}{5}\\ =& 1+0+0+0+-\frac{1}{5}=1-\frac{1}{5}=\frac{4}{5}\end{array}$

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

$\begin{array}{cc}\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})& =(\frac{1}{1}-\frac{1}{1+1})+(\frac{1}{2}-\frac{1}{2+1})+\dots +(\frac{1}{n}-\frac{1}{n+1})\\ & =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots +(\frac{1}{n}-\frac{1}{n+1})\\ & =1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+\dots +(-\frac{1}{n}+\frac{1}{n})-\frac{1}{n+1}\\ & =1+0+0+\dots +0-\frac{1}{n+1}=1-\frac{1}{n+1}\end{array}$

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

Telescoping sum: Definition and explanation (YouTube video by the channel „MJ Education“)

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:

$\begin{array}{cc}\sum _{k=1}^n(a_k-a_{k+1})& =(a_1-a_2)+(a_2-a_3)+\dots +(a_n-a_{n+1})\\ & =a_1+(-a_2+a_2)+(-a_3+a_3)+\dots +(-a_n+a_n)-a_{n+1}\\ & =a_1+0+0+\dots +0-a_{n+1}\\ & =a_1-a_{n+1}\end{array}$

analogously,

$\sum _{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1$

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

Exercise

Prove that $\sum _{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1$.

Solution

There is

$\begin{array}{cc}\sum _{k=1}^n(a_{k+1}-a_k)& =(a_2-a_1)+(a_3-a_2)\dots +(a_n-a_{n-1})+(a_{n+1}-a_n)\\ & =(-a_1+a_2)+(-a_2+a_3)+\dots +(-a_{n-1}+a_n)+(-a_n+a_{n+1})\\ & =-a_1+(-a_2+a_2)+(-a_3+a_3)\dots +(-a_n+a_n)+a_{n+1}\\ & =-a_1+0+0+\dots +0+0+a_{n+1}=a_{n+1}-a_1\end{array}$

Definition and theorem

Definition (telescoping sum)

A telescoping sum is a sum of the form $\sum _{k=1}^n(a_k-a_{k+1})$ or $\sum _{k=1}^n(a_{k+1}-a_k)$.

Theorem (value of a telescoping sum)

There is: $\begin{array}{cc}\sum _{k=1}^n(a_k-a_{k+1})& =a_1-a_{n+1}\\ \sum _{k=1}^n(a_{k+1}-a_k)& =a_{n+1}-a_1\end{array}$

Example (telescoping sum)

Take the sum $\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})$ with $a_k=\frac{1}{k}$ and $a_{k+1}=\frac{1}{k+1}$. We have

$\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})=\frac{1}{1}-\frac{1}{n+1}=1-\frac{1}{n+1}$

Or, putting a $-1$ in front of everything:

$\sum _{k=1}^n(\frac{1}{k+1}-\frac{1}{k})=\frac{1}{n+1}-1$

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:

$\sum _{k=1}^n\frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots +\frac{1}{(n-1)n}+\frac{1}{n(n+1)}$

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:

$\frac{1}{k(k+1)}=\frac{1+(k-k)}{k(k+1)}=\frac{(k+1)-k}{k(k+1)}=\frac{k+1}{k(k+1)}-\frac{k}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$

So

$\sum _{k=1}^n\frac{1}{k(k+1)}=\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})$

And this is a telescoping sum. Who would have guessed that ?! 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

$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k+1)}={\left(\sum _{k=1}^n\frac{1}{k(k+1)}\right)}_{n\in \mathbb{N}}$

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

$\sum _{k=1}^n\frac{1}{k(k+1)}=\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})=1-\frac{1}{n+1}$

So the limit amounts to

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k+1)}& =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n\frac{1}{k(k+1)}\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})\\ & =\underset{n\to \mathrm{\infty }}{lim}(1-\frac{1}{n+1})\\ & \downarrow \text{limit theorems}\\ & =1-\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n+1}=1-0=1\end{array}$

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

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}(a_k-a_{k+1})& ={(\sum _{k=1}^n(a_k-a_{k+1}))}_{n\in \mathbb{N}}\\ & \downarrow \sum _{k=1}^n(a_k-a_{k+1})\text{ is a telescoping sum}\\ & ={(a_1-a_{n+1})}_{n\in \mathbb{N}}\end{array}$

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

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}(a_k-a_{k+1})& =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n(a_k-a_{k+1})\\ & =\underset{n\to \mathrm{\infty }}{lim}(a_1-a_{n+1})\\ & \downarrow \underset{n\to \mathrm{\infty }}{lim}(a_n+b_n)=\underset{n\to \mathrm{\infty }}{lim}{a}_n+\underset{n\to \mathrm{\infty }}{lim}{b}_n\\ & =\underset{n\to \mathrm{\infty }}{lim}{a}_1-\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}=a_1-a\end{array}$

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

$\sum _{k=1}^{\mathrm{\infty }}(a_{k+1}-a_k)=\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}-\underset{n\to \mathrm{\infty }}{lim}{a}_1=a-a_1$

Definition, theorem and example

Definition (telescoping series)

A telescoping series is a series of the form $\sum _{k=1}^{\mathrm{\infty }}(a_k-a_{k+1})$ or $\sum _{k=1}^{\mathrm{\infty }}(a_{k+1}-a_k)$.

Theorem (convergence of telescoping series)

The telescoping series $\sum _{k=1}^{\mathrm{\infty }}(a_k-a_{k+1})$ and $\sum _{k=1}^{\mathrm{\infty }}(a_{k+1}-a_k)$converge if and only if the sequence $(a_n{)}_{n\in \mathbb{N}}$ converges. In that case, the limits are

$\sum _{k=1}^{\mathrm{\infty }}(a_k-a_{k+1})=a_1-\underset{n\to \mathrm{\infty }}{lim}{a}_n$

and

$\sum _{k=1}^{\mathrm{\infty }}(a_{k+1}-a_k)=\underset{n\to \mathrm{\infty }}{lim}{a}_n-a_1$

Example (telescoping series)

The series $\sum _{k=1}^{\mathrm{\infty }}((-1{)}^k-(-1{)}^{k+1})$ diverges, since $a_k=(-1{)}^k$ diverges.

However, the series $\sum _{k=2}^{\mathrm{\infty }}(\sqrt[k]{9}-\sqrt[k+1]{9})$ converges, since the sequence $a_k=\sqrt[k]{9}$ converges to $1$ . The limit of the series is

$\begin{array}{cc}\sum _{k=2}^{\mathrm{\infty }}(\sqrt[k]{9}-\sqrt[k+1]{9})& =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=2}^n(\sqrt[k]{9}-\sqrt[k+1]{9})\\ & \downarrow \text{telescoping sum}\\ & =\underset{n\to \mathrm{\infty }}{lim}(\sqrt[2]{9}-\sqrt[n+1]{9})\\ & =\underset{=3}{\underbrace{\sqrt[2]{9}}}-\underset{=1}{\underbrace{\underset{n\to \mathrm{\infty }}{lim}\sqrt[n+1]{9}}}\\ & =3-1=2\end{array}$

Examples

Example 1

Exercise (Partial sums of the geometric series)

The aim of this exercise is to show the sum formula for geometric series without using induction. What we want to prove is $\sum _{k=0}^n{q}^k=\frac{1-q^{n+1}}{1-q}$ for $q\ne 1$ and $n\in {\mathbb{N}}_0$. Show that the equivalent statement $(1-q)\sum _{k=0}^n{q}^k=1-q^{n+1}$ holds.

Solution (Partial sums of the geometric series)

For $n\in {\mathbb{N}}_0$ and $q\ne 1$ there is

$\begin{array}{cc}(1-q)\sum _{k=0}^n{q}^k& =\sum _{k=0}^n(1-q)\cdot q^k\\ & =\sum _{k=0}^n{q}^k-q^{k+1}\\ & \downarrow \text{telescoping sum}\\ & =q^0-q^{n+1}=1-q^{n+1}\end{array}$

Example 2

Exercise

Does the series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{4k^2-1}$ converge? If yes, determine the limit.

Solution

We need a decomposition of the fraction here, if we want to make it a telescoping series. The denominator can be split in two factors, using the binomial theorems:

$\frac{1}{4k^2-1}=\frac{1}{(2k-1)(2k+1)}$

Now, we can do a partial fraction decomposition as above:

$\begin{array}{cc}\frac{1}{(2k-1)(2k+1)}& =\frac{1+k-k}{(2k-1)(2k+1)}\\ & =\frac{\frac{1}{2}\cdot (2+2k-2k)}{(2k-1)(2k+1)}\\ & =\frac{1}{2}\cdot \frac{(2k+1-2k+1)}{(2k-1)(2k+1)}\\ & =\frac{1}{2}\cdot \frac{(2k+1)-(2k-1)}{(2k-1)(2k+1)}\\ & =\frac{1}{2}\cdot (\frac{2k+1}{(2k-1)(2k+1)}-\frac{2k-1}{(2k-1)(2k+1)})\\ & =\frac{1}{2}\cdot (\frac{1}{2k-1}-\frac{1}{2k+1})\\ \end{array}$

Hence, we get

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{4k^2-1}& =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2}\cdot (\frac{1}{2k-1}-\frac{1}{2k+1})\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}\cdot \sum _{k=1}^n(\frac{1}{2k-1}-\frac{1}{2k+1})\\ & \downarrow \text{telescoping sum}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}\cdot (\frac{1}{2\cdot 1-1}-\frac{1}{2n+1})\\ & \downarrow \underset{n\to \mathrm{\infty }}{lim}\frac{1}{2n+1}=0\\ & =\frac{1}{2}\cdot 1=\frac{1}{2}\end{array}$

Example 3

Exercise

Does the series $\sum _{k=1}^{\mathrm{\infty }}\frac{2k+1}{k(k+1)}$ converge? If yes, determine the limit.

Solution

Again there is only one fraction with a product in the denominator, so we attempt partial fraction decomposition:

$\begin{array}{cc}\frac{2k+1}{k(k+1)}& =\frac{k+k+1}{k(k+1)}\\ & =\frac{k+1}{k(k+1)}+\frac{k}{k(k+1)}\\ & =\frac{1}{k}+\frac{1}{k+1}\\ \end{array}$

This leads us to

$\sum _{k=1}^{\mathrm{\infty }}\frac{2k+1}{k(k+1)}=\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{k}+\frac{1}{k+1})=(1+\frac{1}{2})+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{3}+\frac{1}{4})+\dots$

Be careful: This series is not a telescoping series! We have to add summands - not to subtract them. Even worse, the series does not converge at all: The sequence of partial sums $(S_n{)}_{n\in \mathbb{N}}$ is

$S_n=\sum _{k=1}^n(\frac{1}{k}+\frac{1}{k+1})\ge \sum _{k=1}^n\frac{1}{k}$

So they are greater as for a diverging harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ . By direct comparison,$\sum _{k=1}^{\mathrm{\infty }}\frac{2k+1}{k(k+1)}$ diverges as well. So partial fraction decomposition does not necessarily produce a telescoping sum, but it can be useful to determine whether a series converges or diverges.

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

$a_n=a_1+\sum _{k=2}^n(a_k-a_{k-1})$

Question: Why is there $a_n=a_1+\sum _{k=2}^n(a_k-a_{k-1})$?

There is

$\begin{array}{cc}& a_1+\sum _{k=2}^n(a_k-a_{k-1})\\ & \downarrow \text{index shift}\\ =& a_1+\sum _{k=1}^{n-1}(a_{k+1}-a_k)\\ & \downarrow \text{telescoping sum}\\ =& a_1+(a_n-a_1)\\ =& a_n\end{array}$

So a sequence element can be written as

$a_n=\sum _{k=1}^n{c}_k$ with $c_k=\{\begin{array}{cc}{a}_k-a_{k-1}& ;k\ge 2\\ a_1& ;k=1\end{array}$

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.