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. ^ 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 .