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Telescoping series
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{{Short description|Series whose partial sums eventually only have a fixed number of terms after cancellation}} {{Ref improve|date=March 2021}} In [[mathematics]], a '''telescoping series''' is a [[series (mathematics)|series]] whose general term <math>t_n</math> is of the form <math>t_n=a_{n+1}-a_n</math>, i.e. the difference of two consecutive terms of a [[sequence]] <math>(a_n)</math>. As a consequence the partial sums of the series only consists of two terms of <math>(a_n)</math> after cancellation.<ref name=":0">{{cite book |last1=Apostol |first1=Tom |title=Calculus, Volume 1 |date=1967 |publisher=John Wiley & Sons |edition=Second |pages=386–387 |orig-date=1961}}</ref><ref>Brian S. Thomson and Andrew M. Bruckner, ''Elementary Real Analysis, Second Edition'', CreateSpace, 2008, page 85</ref> The cancellation technique, with part of each term cancelling with part of the next term, is known as the '''method of differences'''. 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''.<ref>{{cite book | last = Weil | first = André | author-link = André Weil | editor1-last = Aubert | editor1-first = Karl Egil | editor1-link = Karl Egil Aubert | editor2-last = Bombieri | editor2-first = Enrico | editor2-link = Enrico Bombieri | editor3-last = Goldfeld | editor3-first = Dorian | editor3-link = Dorian M. Goldfeld | contribution = Prehistory of the zeta-function | doi = 10.1016/B978-0-12-067570-8.50009-3 | location = Boston, Massachusetts | mr = 993308 | pages = 1–9 | publisher = Academic Press | title = Number Theory, Trace Formulas and Discrete Groups: Symposium in Honor of Atle Selberg, Oslo, Norway, July 14–21, 1987 | year = 1989}}</ref> == Definition == [[File:Telescoping Series.png|right|thumb|350px|A telescoping series of powers. Note in the [[summation sign]], <math display="inline">\sum</math>, 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 [[sum (mathematics)|sums]] are finite sums in which pairs of consecutive terms partly cancel each other, leaving only parts of the initial and final terms.<ref name=":0" /><ref>{{cite web |last1=Weisstein |first1=Eric W. |title=Telescoping Sum |url=https://mathworld.wolfram.com/TelescopingSum.html |website=MathWorld |publisher=Wolfram |language=en}}</ref> Let <math>a_n</math> be the elements of a sequence of numbers. Then <math display="block"> \sum_{n=1}^N \left(a_n - a_{n-1}\right) = a_N - a_0.</math> If <math> a_n </math> converges to a limit <math>L</math>, the telescoping [[Series (mathematics)|series]] gives: <math display="block"> \sum_{n=1}^\infty \left(a_n - a_{n-1}\right) = L-a_0. </math> Every series is a telescoping series of its own partial sums.<ref name=":3">{{Cite book |last1=Ablowitz |first1=Mark J. |title=Complex Variables: Introduction and Applications |last2=Fokas |first2=Athanassios S. |publisher=Cambridge University Press |year=2003 |isbn=978-0-521-53429-1 |edition=2nd |pages=110}}</ref> ==Examples== * The product of a [[geometric series]] with initial term <math>a</math> and common ratio <math>r</math> by the factor <math>(1 - r)</math> yields a telescoping sum, which allows for a direct calculation of its limit:<ref>{{cite book |last1=Apostol |first1=Tom |title=Calculus, Volume 1 |date=1967 |publisher=John Wiley & Sons |edition=Second |page=388 |orig-date=1961}}</ref><math display="block">(1 - r) \sum^\infty_{n=0} ar^n = \sum^\infty_{n=0} \left(ar^n - ar^{n+1}\right) = a </math>when <math>|r| < 1,</math> so when <math>|r| < 1,</math> <math display="block"> \sum^\infty_{n=0} ar^n = \frac{a}{1 - r}.</math> * The series<math display="block">\sum_{n=1}^\infty\frac{1}{n(n+1)}</math>is the series of [[Multiplicative inverse|reciprocal]]s of [[pronic number]]s, and it is recognizable as a telescoping series once rewritten in [[Partial fraction decomposition|partial fraction]] form<ref name=":0" /> <math>\begin{align} \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{align}</math> * Let ''k'' be a positive integer. Then<math display="block">\sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k} </math> where ''H''<sub>''k''</sub> is the ''k''th [[harmonic number]]. * Let ''k'' and ''m'' with ''k'' <math>\neq</math> ''m'' be positive integers. Then<math display="block">\sum^\infty_{n=1} {\frac{1}{(n+k)(n+k+1)\dots(n+m-1)(n+m)}} = \frac{1}{m-k} \cdot \frac{k!}{m!} </math> where <math>! </math> denotes the [[factorial]] operation. * Many [[trigonometric function]]s also admit representation as differences, which may reveal telescopic canceling between the consecutive terms. Using the [[angle addition identity]] for a product of sines,<math display="block">\begin{align} \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{align}</math> which does not converge as <math display="inline">N \rightarrow \infty.</math> == Applications == 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 [[memorylessness|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''<sub>''t''</sub> be the number of "occurrences" before time ''t'', and let ''T''<sub>''x''</sub> be the waiting time until the ''x''th "occurrence". We seek the [[probability density function]] of the [[random variable]] ''T''<sub>''x''</sub>. We use the [[probability mass function]] for the Poisson distribution, which tells us that : <math> \Pr(X_t = x) = \frac{(\lambda t)^x e^{-\lambda t}}{x!}, </math> where λ is the average number of occurrences in any time interval of length 1. Observe that the event {''X''<sub>''t''</sub> ≥ x} is the same as the event {''T''<sub>''x''</sub> ≤ ''t''}, and thus they have the same probability. Intuitively, if something occurs at least <math>x</math> times before time <math>t</math>, we have to wait at most <math>t</math> for the <math>xth</math> occurrence. The density function we seek is therefore : <math> \begin{align} f(t) & {} = \frac{d}{dt}\Pr(T_x \le t) = \frac{d}{dt}\Pr(X_t \ge x) = \frac{d}{dt}(1 - \Pr(X_t \le 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^ut^{u-1}}{(u-1)!} - \frac{\lambda^{u+1} t^u}{u!} \right) \end{align} </math> The sum telescopes, leaving : <math> f(t) = \frac{\lambda^x t^{x-1} e^{-\lambda t}}{(x-1)!}. </math> For other applications, see: * [[Proof that the sum of the reciprocals of the primes diverges]], where one of the proofs uses a telescoping sum; * [[Fundamental theorem of calculus]], a continuous analog of telescoping series; * [[Order statistic]], where a telescoping sum occurs in the derivation of a probability density function; * [[Lefschetz fixed-point theorem]], where a telescoping sum arises in [[algebraic topology]]; * [[Homology theory]], again in algebraic topology; * [[Eilenberg–Mazur swindle]], where a telescoping sum of knots occurs; * [[Faddeev–LeVerrier algorithm]]. == Related concepts== A ''telescoping product'' is a finite [[Product (mathematics)|product]] (or the partial product of an infinite product) that can be canceled by the method of quotients to be eventually only a finite number of factors.<ref name="Brilliant">{{cite web |title=Telescoping Series - Product |url=https://brilliant.org/wiki/telescoping-series-product/ |website=Brilliant Math & Science Wiki |publisher=Brilliant.org |access-date=9 February 2020 |language=en-us}}</ref><ref>{{cite web |last1=Bogomolny |first1=Alexander |title=Telescoping Sums, Series and Products |url=https://www.cut-the-knot.org/m/Algebra/TelescopingSums.shtml |website=Cut the Knot |access-date=9 February 2020}}</ref> It is the finite products in which consecutive terms cancel denominator with numerator, leaving only the initial and final terms. Let <math>a_n</math> be a sequence of numbers. Then, <math display="block"> \prod_{n=1}^N \frac{a_{n-1}}{a_n} = \frac{a_0}{a_N}.</math> If <math>a_n </math> converges to 1, the resulting product gives: <math display="block"> \prod_{n=1}^\infty \frac{a_{n-1}}{a_n} = a_0</math> For example, the infinite product<ref name="Brilliant"/> <math display="block">\prod_{n=2}^{\infty} \left(1-\frac{1}{n^2} \right)</math> simplifies as <math display="block">\begin{align} \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}. \end{align}</math> == References == {{reflist}} {{series (mathematics)}} {{DEFAULTSORT:Telescoping Series}} [[Category:Series (mathematics)]]
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