# Telescoping sums and series – Serlo

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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

{\displaystyle {\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.

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

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

{\displaystyle {\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 ${\displaystyle n\in \mathbb {N} }$ of summands:

{\displaystyle {\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 ${\displaystyle \sum _{k=1}^{n}(a_{k}-a_{k+1})}$. Neighbouring terms cancel, so one obtains:

{\displaystyle {\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,

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

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

Exercise

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

Solution

There is

{\displaystyle {\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 ${\displaystyle \sum _{k=1}^{n}(a_{k}-a_{k+1})}$ or ${\displaystyle \sum _{k=1}^{n}(a_{k+1}-a_{k})}$.

Theorem (value of a telescoping sum)

There is:

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

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

Or, putting a ${\displaystyle -1}$ in front of everything:

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

### 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:

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

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 ${\displaystyle k\in \mathbb {N} }$ we have:

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

So

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

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

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

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

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

So the limit amounts to

{\displaystyle {\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 ${\displaystyle \sum _{k=1}^{\infty }(a_{k}-a_{k+1})}$. Their partial sums have the form

{\displaystyle {\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 ${\displaystyle \left(a_{1}-a_{n+1}\right)_{n\in \mathbb {N} }}$ converges. This sequence in turn converges, if and only if ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges. If ${\displaystyle a}$ is the limit of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ , then the limit of the telescoping series amounts to

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

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

### Definition, theorem and example

Definition (telescoping series)

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

Theorem (convergence of telescoping series)

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

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

and

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

Example (telescoping series)

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

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

{\displaystyle {\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 ${\displaystyle \sum _{k=0}^{n}q^{k}={\frac {1-q^{n+1}}{1-q}}}$ for ${\displaystyle q\neq 1}$ and ${\displaystyle n\in \mathbb {N} _{0}}$. Show that the equivalent statement ${\displaystyle (1-q)\sum _{k=0}^{n}q^{k}=1-q^{n+1}}$ holds.

Solution (Partial sums of the geometric series)

For ${\displaystyle n\in \mathbb {N} _{0}}$ and ${\displaystyle q\neq 1}$ there is

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

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

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

{\displaystyle {\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

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

{\displaystyle {\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}}}

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

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 ${\displaystyle (S_{n})_{n\in \mathbb {N} }}$ is

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

So they are greater as for a diverging harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k}}}$ . By direct comparison,${\displaystyle \sum _{k=1}^{\infty }{\tfrac {2k+1}{k(k+1)}}}$ 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 ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ can be made a series if we write it as a telescoping series: We can write

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

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

There is

{\displaystyle {\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

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

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