Telescoping sums and series – Serlo

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

Telescoping sums

Introductory example

Consider the sum

${\begin{aligned}\sum _{k=1}^{4}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)&=\left({\frac {1}{1}}-{\frac {1}{1+1}}\right)+\left({\frac {1}{2}}-{\frac {1}{2+1}}\right)+\left({\frac {1}{3}}-{\frac {1}{3+1}}\right)+\left({\frac {1}{4}}-{\frac {1}{4+1}}\right)\\[0.3em]&=\left({\frac {1}{1}}-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\left({\frac {1}{3}}-{\frac {1}{4}}\right)+\left({\frac {1}{4}}-{\frac {1}{5}}\right)\end{aligned}}$

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.

$\left(1{\color {Red}-{\frac {1}{2}}}\right)+\left({\color {Red}{\frac {1}{2}}}{\color {RedOrange}-{\frac {1}{3}}}\right)+\left({\color {RedOrange}{\frac {1}{3}}}{\color {Olive}-{\frac {1}{4}}}\right)+\left({\color {Olive}{\frac {1}{4}}}{\color {Blue}-{\frac {1}{5}}}\right)$

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

${\begin{aligned}&1+{\color {Red}\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)}+{\color {RedOrange}\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)}+{\color {Olive}\left(-{\frac {1}{4}}+{\frac {1}{4}}\right)}-{\frac {1}{5}}\\[0.3em]=&1+{\color {Red}0}+{\color {RedOrange}0}+{\color {Olive}0}+-{\frac {1}{5}}=1-{\frac {1}{5}}={\frac {4}{5}}\end{aligned}}$

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

${\begin{aligned}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)&=\left({\frac {1}{1}}-{\frac {1}{1+1}}\right)+\left({\frac {1}{2}}-{\frac {1}{2+1}}\right)+\ldots +\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\[0.3em]&=\left(1{\color {Red}-{\frac {1}{2}}}\right)+\left({\color {Red}{\frac {1}{2}}}{\color {RedOrange}-{\frac {1}{3}}}\right)+\ldots +\left({\color {Purple}{\frac {1}{n}}}-{\frac {1}{n+1}}\right)\\[0.3em]&=1+{\color {Red}\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)}+{\color {RedOrange}\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)}+\ldots +{\color {Purple}\left(-{\frac {1}{n}}+{\frac {1}{n}}\right)}-{\frac {1}{n+1}}\\[0.3em]&=1+{\color {Red}0}+{\color {RedOrange}0}+\ldots +{\color {Purple}0}-{\frac {1}{n+1}}=1-{\frac {1}{n+1}}\end{aligned}}$

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 collapsible 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:

${\begin{aligned}\sum _{k=1}^{n}(a_{k}-a_{k+1})&=(a_{1}{\color {Red}-a_{2}})+({\color {Red}a_{2}}{\color {RedOrange}-a_{3}})+\ldots +({\color {Purple}a_{n}}-a_{n+1})\\&=a_{1}+{\color {Red}(-a_{2}+a_{2})}+{\color {RedOrange}(-a_{3}+a_{3})}+\ldots +{\color {Purple}(-a_{n}+a_{n})}-a_{n+1}\\&=a_{1}+{\color {Red}0}+{\color {RedOrange}0}+\ldots +{\color {Purple}0}-a_{n+1}\\&=a_{1}-a_{n+1}\end{aligned}}$

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

${\begin{aligned}\sum _{k=1}^{n}(a_{k+1}-a_{k})&=({\color {Red}a_{2}}-a_{1})+({\color {RedOrange}a_{3}}{\color {Red}-a_{2}})\ldots +({\color {Purple}a_{n}}{\color {Blue}-a_{n-1}})+(a_{n+1}{\color {Purple}-a_{n}})\\&=(-a_{1}{\color {Red}+a_{2}})+({\color {Red}-a_{2}}{\color {RedOrange}+a_{3}})+\ldots +({\color {Blue}-a_{n-1}}{\color {Purple}+a_{n}})+({\color {Purple}-a_{n}}+a_{n+1})\\&=-a_{1}+{\color {Red}(-a_{2}+a_{2})}+{\color {RedOrange}(-a_{3}+a_{3})}\ldots +{\color {Purple}(-a_{n}+a_{n})}+a_{n+1}\\&=-a_{1}+{\color {Red}0}+{\color {RedOrange}0}+\ldots +{\color {Blue}0}+{\color {Purple}0}+a_{n+1}=a_{n+1}-a_{1}\end{aligned}}$

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:

${\begin{aligned}\sum _{k=1}^{n}(a_{k}-a_{k+1})&=a_{1}-a_{n+1}\\[0.5em]\sum _{k=1}^{n}(a_{k+1}-a_{k})&=a_{n+1}-a_{1}\end{aligned}}$

Example (telescoping sum)

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

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

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

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

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:

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

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

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

And this is a telescoping sum. Who would have guessed that ?!   The re-formulation ${\tfrac {1}{k(k+1)}}={\tfrac {1}{k}}-{\tfrac {1}{k+1}}$ 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

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

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

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

So the limit amounts to

${\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}&=\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)}}\\[0.3em]&=\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)\\[0.3em]&=\lim _{n\to \infty }\left(1-{\frac {1}{n+1}}\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{limit theorems}}\right.}\\[0.3em]&=1-\lim _{n\to \infty }{\frac {1}{n+1}}=1-0=1\end{aligned}}$

General introduction

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

${\begin{aligned}\sum _{k=1}^{\infty }(a_{k}-a_{k+1})&=\left(\sum _{k=1}^{n}(a_{k}-a_{k+1})\right)_{n\in \mathbb {N} }\\[1em]&{\color {OliveGreen}\left\downarrow \ \sum _{k=1}^{n}(a_{k}-a_{k+1}){\text{ is a telescoping sum}}\right.}\\[1em]&=\left(a_{1}-a_{n+1}\right)_{n\in \mathbb {N} }\end{aligned}}$

To see whether a telescoping series converges, we need to investigate whether the sequence $\left(a_{1}-a_{n+1}\right)_{n\in \mathbb {N} }$ 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

${\begin{aligned}\sum _{k=1}^{\infty }(a_{k}-a_{k+1})&=\lim _{n\to \infty }\sum _{k=1}^{n}(a_{k}-a_{k+1})\\[1em]&=\lim _{n\to \infty }(a_{1}-a_{n+1})\\[1em]&{\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }(a_{n}+b_{n})=\lim _{n\to \infty }a_{n}+\lim _{n\to \infty }b_{n}\right.}\\[1em]&=\lim _{n\to \infty }a_{1}-\lim _{n\to \infty }a_{n+1}=a_{1}-a\end{aligned}}$

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 $\sum _{k=1}^{\infty }(a_{k+1}-a_{k})=(a_{n+1}-a_{1})_{n\in \mathbb {N} }$ converges, if we can show that $(a_{n})_{n\in \mathbb {N} }$ converges. In that case, the limit is

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

Definition, theorem and example

Definition (telescoping series)

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

Theorem (convergence of telescoping series)

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

$\sum _{k=1}^{\infty }(a_{k}-a_{k+1})=a_{1}-\lim _{n\to \infty }a_{n}$

and

$\sum _{k=1}^{\infty }(a_{k+1}-a_{k})=\lim _{n\to \infty }a_{n}-a_{1}$

Example (telescoping series)

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

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

${\begin{aligned}\sum _{k=2}^{\infty }\left({\sqrt[{k}]{9}}-{\sqrt[{k+1}]{9}}\right)&=\lim _{n\to \infty }\sum _{k=2}^{n}\left({\sqrt[{k}]{9}}-{\sqrt[{k+1}]{9}}\right)\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{telescoping sum}}\right.}\\[1em]&=\lim _{n\to \infty }\left({\sqrt[{2}]{9}}-{\sqrt[{n+1}]{9}}\right)\\[1em]&=\underbrace {\sqrt[{2}]{9}} _{=\ 3}-\underbrace {\lim _{n\to \infty }{\sqrt[{n+1}]{9}}} _{=1}\\[1em]&=3-1=2\end{aligned}}$

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 $q\neq 1$ 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 $q\neq 1$ there is

${\begin{aligned}(1-q)\sum _{k=0}^{n}q^{k}&=\sum _{k=0}^{n}(1-q)\cdot q^{k}\\[0.3em]&=\sum _{k=0}^{n}q^{k}-q^{k+1}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{telescoping sum}}\right.}\\[0.3em]&=q^{0}-q^{n+1}=1-q^{n+1}\end{aligned}}$

Example 2

Exercise

Does the series $\sum _{k=1}^{\infty }{\tfrac {1}{4k^{2}-1}}$ 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:

${\begin{aligned}{\frac {1}{(2k-1)(2k+1)}}&={\frac {1+k-k}{(2k-1)(2k+1)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (2+2k-2k)}{(2k-1)(2k+1)}}\\[0.5em]&={\frac {1}{2}}\cdot {\frac {(2k+1-2k+1)}{(2k-1)(2k+1)}}\\[0.5em]&={\frac {1}{2}}\cdot {\frac {(2k+1)-(2k-1)}{(2k-1)(2k+1)}}\\[0.5em]&={\frac {1}{2}}\cdot \left({\frac {2k+1}{(2k-1)(2k+1)}}-{\frac {2k-1}{(2k-1)(2k+1)}}\right)\\[0.5em]&={\frac {1}{2}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+1}}\right)\\[0.5em]\end{aligned}}$

Hence, we get

${\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{4k^{2}-1}}&=\sum _{k=1}^{\infty }{\frac {1}{2}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+1}}\right)\\[1em]&=\lim _{n\to \infty }{\frac {1}{2}}\cdot \sum _{k=1}^{n}\left({\frac {1}{2k-1}}-{\frac {1}{2k+1}}\right)\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{telescoping sum}}\right.}\\[1em]&=\lim _{n\to \infty }{\frac {1}{2}}\cdot \left({\frac {1}{2\cdot 1-1}}-{\frac {1}{2n+1}}\right)\\[1em]&{\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }{\frac {1}{2n+1}}=0\right.}\\[1em]&={\frac {1}{2}}\cdot 1={\frac {1}{2}}\end{aligned}}$

Example 3

Exercise

Does the series $\sum _{k=1}^{\infty }{\tfrac {2k+1}{k(k+1)}}$ 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:

${\begin{aligned}{\frac {2k+1}{k(k+1)}}&={\frac {k+k+1}{k(k+1)}}\\[0.5em]&={\frac {k+1}{k(k+1)}}+{\frac {k}{k(k+1)}}\\[0.5em]&={\frac {1}{k}}+{\frac {1}{k+1}}\\[0.5em]\end{aligned}}$

This leads us to

$\sum _{k=1}^{\infty }{\frac {2k+1}{k(k+1)}}=\sum _{k=1}^{\infty }\left({\frac {1}{k}}+{\frac {1}{k+1}}\right)=\left(1+{\frac {1}{2}}\right)+\left({\frac {1}{2}}+{\frac {1}{3}}\right)+\left({\frac {1}{3}}+{\frac {1}{4}}\right)+\ldots$

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

$S_{n}=\sum _{k=1}^{n}\left({\frac {1}{k}}+{\frac {1}{k+1}}\right)\geq \sum _{k=1}^{n}{\frac {1}{k}}$

So they are greater as for a diverging harmonic series $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$ . By direct comparison,$\sum _{k=1}^{\infty }{\tfrac {2k+1}{k(k+1)}}$ 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

${\begin{aligned}&a_{1}+\sum _{k=2}^{n}(a_{k}-a_{k-1})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{index shift}}\right.}\\[0.3em]=\ &a_{1}+\sum _{k=1}^{n-1}(a_{k+1}-a_{k})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{telescoping sum}}\right.}\\[0.3em]=\ &a_{1}+(a_{n}-a_{1})\\[0.3em]=\ &a_{n}\end{aligned}}$

So a sequence element can be written as

$a_{n}=\sum _{k=1}^{n}c_{k}$ with $c_{k}={\begin{cases}a_{k}-a_{k-1}&;k\geq 2\\a_{1}&;k=1\end{cases}}$

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