Template:Short description Template:Ref improve In mathematics, a telescoping series is a 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">Template:Cite book</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>Template:Cite book</ref>

DefinitionEdit

File:Telescoping Series.png
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 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>{{#invoke:citation/CS1|citation |CitationClass=web }}</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 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">Template:Cite book</ref>

ExamplesEdit

  • 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>Template:Cite book</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 reciprocals of pronic numbers, and it is recognizable as a telescoping series once rewritten in 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 Hk is the kth 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 functions 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>

ApplicationsEdit

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 Xt be the number of "occurrences" before time t, and let Tx be the waiting time until the xth "occurrence". We seek the probability density function of the random variable Tx. 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 {Xt ≥ x} is the same as the event {Txt}, 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:

Related conceptsEdit

A telescoping product is a finite 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">{{#invoke:citation/CS1|citation |CitationClass=web }}</ref><ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</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>

ReferencesEdit

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