Telescoping series are certain series where summands cancel against each other. This makes evaluating them particularly easy.
Consider the sum
∑
k
=
1
4
(
1
k
−
1
k
+
1
)
=
(
1
1
−
1
1
+
1
)
+
(
1
2
−
1
2
+
1
)
+
(
1
3
−
1
3
+
1
)
+
(
1
4
−
1
4
+
1
)
=
(
1
1
−
1
2
)
+
(
1
2
−
1
3
)
+
(
1
3
−
1
4
)
+
(
1
4
−
1
5
)
{\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.
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
(
1
3
−
1
4
)
+
(
1
4
−
1
5
)
{\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
1
+
(
−
1
2
+
1
2
)
+
(
−
1
3
+
1
3
)
+
(
−
1
4
+
1
4
)
−
1
5
=
1
+
0
+
0
+
0
+
−
1
5
=
1
−
1
5
=
4
5
{\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
n
∈
N
{\displaystyle n\in \mathbb {N} }
of summands:
∑
k
=
1
n
(
1
k
−
1
k
+
1
)
=
(
1
1
−
1
1
+
1
)
+
(
1
2
−
1
2
+
1
)
+
…
+
(
1
n
−
1
n
+
1
)
=
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
…
+
(
1
n
−
1
n
+
1
)
=
1
+
(
−
1
2
+
1
2
)
+
(
−
1
3
+
1
3
)
+
…
+
(
−
1
n
+
1
n
)
−
1
n
+
1
=
1
+
0
+
0
+
…
+
0
−
1
n
+
1
=
1
−
1
n
+
1
{\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.
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
∑
k
=
1
n
(
a
k
−
a
k
+
1
)
{\displaystyle \sum _{k=1}^{n}(a_{k}-a_{k+1})}
. Neighbouring terms cancel, so one obtains:
∑
k
=
1
n
(
a
k
−
a
k
+
1
)
=
(
a
1
−
a
2
)
+
(
a
2
−
a
3
)
+
…
+
(
a
n
−
a
n
+
1
)
=
a
1
+
(
−
a
2
+
a
2
)
+
(
−
a
3
+
a
3
)
+
…
+
(
−
a
n
+
a
n
)
−
a
n
+
1
=
a
1
+
0
+
0
+
…
+
0
−
a
n
+
1
=
a
1
−
a
n
+
1
{\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,
∑
k
=
1
n
(
a
k
+
1
−
a
k
)
=
a
n
+
1
−
a
1
{\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
∑
k
=
1
n
(
a
k
+
1
−
a
k
)
=
a
n
+
1
−
a
1
{\displaystyle \sum _{k=1}^{n}(a_{k+1}-a_{k})=a_{n+1}-a_{1}}
.
Solution
There is
∑
k
=
1
n
(
a
k
+
1
−
a
k
)
=
(
a
2
−
a
1
)
+
(
a
3
−
a
2
)
…
+
(
a
n
−
a
n
−
1
)
+
(
a
n
+
1
−
a
n
)
=
(
−
a
1
+
a
2
)
+
(
−
a
2
+
a
3
)
+
…
+
(
−
a
n
−
1
+
a
n
)
+
(
−
a
n
+
a
n
+
1
)
=
−
a
1
+
(
−
a
2
+
a
2
)
+
(
−
a
3
+
a
3
)
…
+
(
−
a
n
+
a
n
)
+
a
n
+
1
=
−
a
1
+
0
+
0
+
…
+
0
+
0
+
a
n
+
1
=
a
n
+
1
−
a
1
{\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 (telescoping sum)
A telescoping sum is a sum of the form
∑
k
=
1
n
(
a
k
−
a
k
+
1
)
{\displaystyle \sum _{k=1}^{n}(a_{k}-a_{k+1})}
or
∑
k
=
1
n
(
a
k
+
1
−
a
k
)
{\displaystyle \sum _{k=1}^{n}(a_{k+1}-a_{k})}
.
Theorem (value of a telescoping sum)
There is:
∑
k
=
1
n
(
a
k
−
a
k
+
1
)
=
a
1
−
a
n
+
1
∑
k
=
1
n
(
a
k
+
1
−
a
k
)
=
a
n
+
1
−
a
1
{\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}}}
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:
∑
k
=
1
n
1
k
(
k
+
1
)
=
1
1
⋅
2
+
1
2
⋅
3
+
1
3
⋅
4
+
…
+
1
(
n
−
1
)
n
+
1
n
(
n
+
1
)
{\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
k
∈
N
{\displaystyle k\in \mathbb {N} }
we have:
1
k
(
k
+
1
)
=
1
+
(
k
−
k
)
k
(
k
+
1
)
=
(
k
+
1
)
−
k
k
(
k
+
1
)
=
k
+
1
k
(
k
+
1
)
−
k
k
(
k
+
1
)
=
1
k
−
1
k
+
1
{\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
∑
k
=
1
n
1
k
(
k
+
1
)
=
∑
k
=
1
n
(
1
k
−
1
k
+
1
)
{\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
1
k
(
k
+
1
)
=
1
k
−
1
k
+
1
{\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.
Solution (Partial sums of the geometric series)
For
n
∈
N
0
{\displaystyle n\in \mathbb {N} _{0}}
and
q
≠
1
{\displaystyle q\neq 1}
there is
(
1
−
q
)
∑
k
=
0
n
q
k
=
∑
k
=
0
n
(
1
−
q
)
⋅
q
k
=
∑
k
=
0
n
q
k
−
q
k
+
1
↓
telescoping sum
=
q
0
−
q
n
+
1
=
1
−
q
n
+
1
{\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}}}
Exercise
Does the series
∑
k
=
1
∞
1
4
k
2
−
1
{\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:
1
4
k
2
−
1
=
1
(
2
k
−
1
)
(
2
k
+
1
)
{\displaystyle {\frac {1}{4k^{2}-1}}={\frac {1}{(2k-1)(2k+1)}}}
Now, we can do a partial fraction decomposition as above:
1
(
2
k
−
1
)
(
2
k
+
1
)
=
1
+
k
−
k
(
2
k
−
1
)
(
2
k
+
1
)
=
1
2
⋅
(
2
+
2
k
−
2
k
)
(
2
k
−
1
)
(
2
k
+
1
)
=
1
2
⋅
(
2
k
+
1
−
2
k
+
1
)
(
2
k
−
1
)
(
2
k
+
1
)
=
1
2
⋅
(
2
k
+
1
)
−
(
2
k
−
1
)
(
2
k
−
1
)
(
2
k
+
1
)
=
1
2
⋅
(
2
k
+
1
(
2
k
−
1
)
(
2
k
+
1
)
−
2
k
−
1
(
2
k
−
1
)
(
2
k
+
1
)
)
=
1
2
⋅
(
1
2
k
−
1
−
1
2
k
+
1
)
{\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
∑
k
=
1
∞
1
4
k
2
−
1
=
∑
k
=
1
∞
1
2
⋅
(
1
2
k
−
1
−
1
2
k
+
1
)
=
lim
n
→
∞
1
2
⋅
∑
k
=
1
n
(
1
2
k
−
1
−
1
2
k
+
1
)
↓
telescoping sum
=
lim
n
→
∞
1
2
⋅
(
1
2
⋅
1
−
1
−
1
2
n
+
1
)
↓
lim
n
→
∞
1
2
n
+
1
=
0
=
1
2
⋅
1
=
1
2
{\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}}}
Exercise
Does the series
∑
k
=
1
∞
2
k
+
1
k
(
k
+
1
)
{\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:
2
k
+
1
k
(
k
+
1
)
=
k
+
k
+
1
k
(
k
+
1
)
=
k
+
1
k
(
k
+
1
)
+
k
k
(
k
+
1
)
=
1
k
+
1
k
+
1
{\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}}}
This leads us to
∑
k
=
1
∞
2
k
+
1
k
(
k
+
1
)
=
∑
k
=
1
∞
(
1
k
+
1
k
+
1
)
=
(
1
+
1
2
)
+
(
1
2
+
1
3
)
+
(
1
3
+
1
4
)
+
…
{\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
(
S
n
)
n
∈
N
{\displaystyle (S_{n})_{n\in \mathbb {N} }}
is
S
n
=
∑
k
=
1
n
(
1
k
+
1
k
+
1
)
≥
∑
k
=
1
n
1
k
{\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
∑
k
=
1
∞
1
k
{\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k}}}
. By direct comparison,
∑
k
=
1
∞
2
k
+
1
k
(
k
+
1
)
{\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.
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
∈
N
{\displaystyle (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
+
∑
k
=
2
n
(
a
k
−
a
k
−
1
)
{\displaystyle a_{n}=a_{1}+\sum _{k=2}^{n}(a_{k}-a_{k-1})}
Question: Why is there
a
n
=
a
1
+
∑
k
=
2
n
(
a
k
−
a
k
−
1
)
{\displaystyle a_{n}=a_{1}+\sum _{k=2}^{n}(a_{k}-a_{k-1})}
?
There is
a
1
+
∑
k
=
2
n
(
a
k
−
a
k
−
1
)
↓
index shift
=
a
1
+
∑
k
=
1
n
−
1
(
a
k
+
1
−
a
k
)
↓
telescoping sum
=
a
1
+
(
a
n
−
a
1
)
=
a
n
{\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
a
n
=
∑
k
=
1
n
c
k
{\displaystyle a_{n}=\sum _{k=1}^{n}c_{k}}
with
c
k
=
{
a
k
−
a
k
−
1
;
k
≥
2
a
1
;
k
=
1
{\displaystyle c_{k}={\begin{cases}a_{k}-a_{k-1}&;k\geq 2\\a_{1}&;k=1\end{cases}}}
The sequence
(
a
n
)
n
∈
N
{\displaystyle (a_{n})_{n\in \mathbb {N} }}
can hence be interpreted as a series
∑
k
=
1
∞
c
k
{\displaystyle \sum _{k=1}^{\infty }c_{k}}
, where the "series" is seen identical with "sequence of partial sums", here.