Series whose partial sums eventually only have a fixed number of terms after cancellation
In mathematics , a telescoping series is a series whose general term
t
n
{\displaystyle t_{n}}
is of the form
t
n
=
a
n
+
1
−
a
n
{\displaystyle t_{n}=a_{n+1}-a_{n}}
, i.e. the difference of two consecutive terms of a sequence
(
a
n
)
{\displaystyle (a_{n})}
.[1]
As a consequence the partial sums only consists of two terms of
(
a
n
)
{\displaystyle (a_{n})}
after cancellation.[2] [3] The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences .
For example, the series
∑
n
=
1
∞
1
n
(
n
+
1
)
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}}
(the series of reciprocals of pronic numbers ) simplifies as
∑
n
=
1
∞
1
n
(
n
+
1
)
=
∑
n
=
1
∞
(
1
n
−
1
n
+
1
)
=
lim
N
→
∞
∑
n
=
1
N
(
1
n
−
1
n
+
1
)
=
lim
N
→
∞
[
(
1
−
1
2
)
+
(
1
2
−
1
3
)
+
⋯
+
(
1
N
−
1
N
+
1
)
]
=
lim
N
→
∞
[
1
+
(
−
1
2
+
1
2
)
+
(
−
1
3
+
1
3
)
+
⋯
+
(
−
1
N
+
1
N
)
−
1
N
+
1
]
=
lim
N
→
∞
[
1
−
1
N
+
1
]
=
1.
{\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}&{}=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\sum _{n=1}^{N}\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\left\lbrack {\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1-{\frac {1}{N+1}}}\right\rbrack =1.\end{aligned}}}
An early statement of the formula for the sum or partial sums of a telescoping series can be found in a 1644 work by Evangelista Torricelli , De dimensione parabolae .[4]
In general [ edit ]
A telescoping series of powers. Note in the summation sign ,
∑
{\textstyle \sum }
, the index n goes from 1 to m . There is no relationship between n and m beyond the fact that both are natural numbers .
Telescoping sums are finite sums in which pairs of consecutive terms cancel each other, leaving only the initial and final terms.[5]
Let
a
n
{\displaystyle a_{n}}
be a sequence of numbers. Then,
∑
n
=
1
N
(
a
n
−
a
n
−
1
)
=
a
N
−
a
0
{\displaystyle \sum _{n=1}^{N}\left(a_{n}-a_{n-1}\right)=a_{N}-a_{0}}
If
a
n
→
0
{\displaystyle a_{n}\rightarrow 0}
∑
n
=
1
∞
(
a
n
−
a
n
−
1
)
=
−
a
0
{\displaystyle \sum _{n=1}^{\infty }\left(a_{n}-a_{n-1}\right)=-a_{0}}
Telescoping products are finite products in which consecutive terms cancel denominator with numerator, leaving only the initial and final terms.
Let
a
n
{\displaystyle a_{n}}
be a sequence of numbers. Then,
∏
n
=
1
N
a
n
−
1
a
n
=
a
0
a
N
{\displaystyle \prod _{n=1}^{N}{\frac {a_{n-1}}{a_{n}}}={\frac {a_{0}}{a_{N}}}}
If
a
n
→
1
{\displaystyle a_{n}\rightarrow 1}
∏
n
=
1
∞
a
n
−
1
a
n
=
a
0
{\displaystyle \prod _{n=1}^{\infty }{\frac {a_{n-1}}{a_{n}}}=a_{0}}
More examples [ edit ]
Many trigonometric functions also admit representation as a difference, which allows telescopic canceling between the consecutive terms.
∑
n
=
1
N
sin
(
n
)
=
∑
n
=
1
N
1
2
csc
(
1
2
)
(
2
sin
(
1
2
)
sin
(
n
)
)
=
1
2
csc
(
1
2
)
∑
n
=
1
N
(
cos
(
2
n
−
1
2
)
−
cos
(
2
n
+
1
2
)
)
=
1
2
csc
(
1
2
)
(
cos
(
1
2
)
−
cos
(
2
N
+
1
2
)
)
.
{\displaystyle {\begin{aligned}\sum _{n=1}^{N}\sin \left(n\right)&{}=\sum _{n=1}^{N}{\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left(2\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\right)\\&{}={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\sum _{n=1}^{N}\left(\cos \left({\frac {2n-1}{2}}\right)-\cos \left({\frac {2n+1}{2}}\right)\right)\\&{}={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left(\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+1}{2}}\right)\right).\end{aligned}}}
Some sums of the form
∑
n
=
1
N
f
(
n
)
g
(
n
)
{\displaystyle \sum _{n=1}^{N}{f(n) \over g(n)}}
where f and g are polynomial functions whose quotient may be broken up into partial fractions , will fail to admit summation by this method. In particular, one has
∑
n
=
0
∞
2
n
+
3
(
n
+
1
)
(
n
+
2
)
=
∑
n
=
0
∞
(
1
n
+
1
+
1
n
+
2
)
=
(
1
1
+
1
2
)
+
(
1
2
+
1
3
)
+
(
1
3
+
1
4
)
+
⋯
⋯
+
(
1
n
−
1
+
1
n
)
+
(
1
n
+
1
n
+
1
)
+
(
1
n
+
1
+
1
n
+
2
)
+
⋯
=
∞
.
{\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }{\frac {2n+3}{(n+1)(n+2)}}={}&\sum _{n=0}^{\infty }\left({\frac {1}{n+1}}+{\frac {1}{n+2}}\right)\\={}&\left({\frac {1}{1}}+{\frac {1}{2}}\right)+\left({\frac {1}{2}}+{\frac {1}{3}}\right)+\left({\frac {1}{3}}+{\frac {1}{4}}\right)+\cdots \\&{}\cdots +\left({\frac {1}{n-1}}+{\frac {1}{n}}\right)+\left({\frac {1}{n}}+{\frac {1}{n+1}}\right)+\left({\frac {1}{n+1}}+{\frac {1}{n+2}}\right)+\cdots \\={}&\infty .\end{aligned}}}
The problem is that the terms do not cancel.
Let k be a positive integer. Then
∑
n
=
1
∞
1
n
(
n
+
k
)
=
H
k
k
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+k)}}={\frac {H_{k}}{k}}}
where H k is the k th harmonic number . All of the terms after 1/(k − 1) cancel.
Let k,m with k
≠
{\displaystyle \neq }
m be positive integers. Then
∑
n
=
1
∞
1
(
n
+
k
)
(
n
+
k
+
1
)
…
(
n
+
m
−
1
)
(
n
+
m
)
=
1
m
−
k
⋅
k
!
m
!
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(n+k)(n+k+1)\dots (n+m-1)(n+m)}}={\frac {1}{m-k}}\cdot {\frac {k!}{m!}}}
An application in probability theory [ edit ]
In probability theory , a Poisson process is a stochastic process of which the simplest case involves "occurrences" at random times, the waiting time until the next occurrence having a memoryless exponential distribution , and the number of "occurrences" in any time interval having a Poisson distribution whose expected value is proportional to the length of the time interval. Let X t be the number of "occurrences" before time t , and let T x be the waiting time until the x th "occurrence". We seek the probability density function of the random variable T x . We use the probability mass function for the Poisson distribution, which tells us that
Pr
(
X
t
=
x
)
=
(
λ
t
)
x
e
−
λ
t
x
!
,
{\displaystyle \Pr(X_{t}=x)={\frac {(\lambda t)^{x}e^{-\lambda t}}{x!}},}
where λ is the average number of occurrences in any time interval of length 1. Observe that the event {X t ≥ x} is the same as the event {T x ≤ t }, and thus they have the same probability. Intuitively, if something occurs at least
x
{\displaystyle x}
times before time
t
{\displaystyle t}
, we have to wait at most
t
{\displaystyle t}
for the
x
t
h
{\displaystyle xth}
occurrence. The density function we seek is therefore
f
(
t
)
=
d
d
t
Pr
(
T
x
≤
t
)
=
d
d
t
Pr
(
X
t
≥
x
)
=
d
d
t
(
1
−
Pr
(
X
t
≤
x
−
1
)
)
=
d
d
t
(
1
−
∑
u
=
0
x
−
1
Pr
(
X
t
=
u
)
)
=
d
d
t
(
1
−
∑
u
=
0
x
−
1
(
λ
t
)
u
e
−
λ
t
u
!
)
=
λ
e
−
λ
t
−
e
−
λ
t
∑
u
=
1
x
−
1
(
λ
u
t
u
−
1
(
u
−
1
)
!
−
λ
u
+
1
t
u
u
!
)
{\displaystyle {\begin{aligned}f(t)&{}={\frac {d}{dt}}\Pr(T_{x}\leq t)={\frac {d}{dt}}\Pr(X_{t}\geq x)={\frac {d}{dt}}(1-\Pr(X_{t}\leq x-1))\\\\&{}={\frac {d}{dt}}\left(1-\sum _{u=0}^{x-1}\Pr(X_{t}=u)\right)={\frac {d}{dt}}\left(1-\sum _{u=0}^{x-1}{\frac {(\lambda t)^{u}e^{-\lambda t}}{u!}}\right)\\\\&{}=\lambda e^{-\lambda t}-e^{-\lambda t}\sum _{u=1}^{x-1}\left({\frac {\lambda ^{u}t^{u-1}}{(u-1)!}}-{\frac {\lambda ^{u+1}t^{u}}{u!}}\right)\end{aligned}}}
The sum telescopes, leaving
f
(
t
)
=
λ
x
t
x
−
1
e
−
λ
t
(
x
−
1
)
!
.
{\displaystyle f(t)={\frac {\lambda ^{x}t^{x-1}e^{-\lambda t}}{(x-1)!}}.}
Similar concepts [ edit ]
Telescoping series [ edit ]
A telescoping product is a finite product (or the partial product of an infinite product) that can be cancelled by method of quotients to be eventually only a finite number of factors.[6] [7]
For example, the infinite product[6]
∏
n
=
2
∞
(
1
−
1
n
2
)
{\displaystyle \prod _{n=2}^{\infty }\left(1-{\frac {1}{n^{2}}}\right)}
simplifies as
∏
n
=
2
∞
(
1
−
1
n
2
)
=
∏
n
=
2
∞
(
n
−
1
)
(
n
+
1
)
n
2
=
lim
N
→
∞
∏
n
=
2
N
n
−
1
n
×
∏
n
=
2
N
n
+
1
n
=
lim
N
→
∞
[
1
2
×
2
3
×
3
4
×
⋯
×
N
−
1
N
]
×
[
3
2
×
4
3
×
5
4
×
⋯
×
N
N
−
1
×
N
+
1
N
]
=
lim
N
→
∞
[
1
2
]
×
[
N
+
1
N
]
=
1
2
×
lim
N
→
∞
[
N
+
1
N
]
=
1
2
×
lim
N
→
∞
[
N
N
+
1
N
]
=
1
2
.
{\displaystyle {\begin{aligned}\prod _{n=2}^{\infty }\left(1-{\frac {1}{n^{2}}}\right)&=\prod _{n=2}^{\infty }{\frac {(n-1)(n+1)}{n^{2}}}\\&=\lim _{N\to \infty }\prod _{n=2}^{N}{\frac {n-1}{n}}\times \prod _{n=2}^{N}{\frac {n+1}{n}}\\&=\lim _{N\to \infty }\left\lbrack {{\frac {1}{2}}\times {\frac {2}{3}}\times {\frac {3}{4}}\times \cdots \times {\frac {N-1}{N}}}\right\rbrack \times \left\lbrack {{\frac {3}{2}}\times {\frac {4}{3}}\times {\frac {5}{4}}\times \cdots \times {\frac {N}{N-1}}\times {\frac {N+1}{N}}}\right\rbrack \\&=\lim _{N\to \infty }\left\lbrack {\frac {1}{2}}\right\rbrack \times \left\lbrack {\frac {N+1}{N}}\right\rbrack \\&={\frac {1}{2}}\times \lim _{N\to \infty }\left\lbrack {\frac {N+1}{N}}\right\rbrack \\&={\frac {1}{2}}\times \lim _{N\to \infty }\left\lbrack {\frac {N}{N}}+{\frac {1}{N}}\right\rbrack \\&={\frac {1}{2}}.\end{aligned}}}
Other applications [ edit ]
For other applications, see:
References [ edit ]
^ Apostol, Tom (1967). Calculus, Volume 1 (Second ed.). John Wiley & Sons. p. 386.
^ Tom M. Apostol , Calculus, Volume 1, Blaisdell Publishing Company, 1962, pages 422–3
^ Brian S. Thomson and Andrew M. Bruckner, Elementary Real Analysis, Second Edition , CreateSpace, 2008, page 85
^ Weil, André (1989). "Prehistory of the zeta-function". In Aubert, Karl Egil ; Bombieri, Enrico ; Goldfeld, Dorian (eds.). Number Theory, Trace Formulas and Discrete Groups: Symposium in Honor of Atle Selberg, Oslo, Norway, July 14–21, 1987 . Boston, Massachusetts: Academic Press. pp. 1–9. doi :10.1016/B978-0-12-067570-8.50009-3 . MR 0993308 .
^ Weisstein, Eric W. "Telescoping Sum" . MathWorld . Wolfram.
^ Jump up to: a b "Telescoping Series - Product" . Brilliant Math & Science Wiki . Brilliant.org. Retrieved 9 February 2020 .
^ Bogomolny, Alexander. "Telescoping Sums, Series and Products" . Cut the Knot . Retrieved 9 February 2020 .