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| In [[mathematics]], a '''telescoping series''' is a [[series (mathematics)|series]] whose partial sums eventually only have a fixed number of terms after cancellation.<ref>[[Tom M. Apostol]], ''Calculus, Volume 1,'' Blaisdell Publishing Company, 1962, pages 422–3</ref><ref>Brian S. Thomson and Andrew M. Bruckner, ''Elementary Real Analysis, Second Edition'', CreateSpace, 2008, page 85</ref> Such a technique is also known as the '''method of differences'''.
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| For example, the series
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| :<math>\sum_{n=1}^\infty\frac{1}{n(n+1)}</math>
| | His name is Merlin but it is not the most masucline name out that there. Guam has for ages been her home but now she is considering likewise. My husband doesn't like it the way I do but the things i really look foward to is driving but I struggle to search out time for this. Production and planning precisely what she executes. Check out one of the most news website: http://www.quantumpendants.org/<br><br>My web site: [http://www.quantumpendants.org/ Quantum Energy Pendant Benefits] |
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| simplifies as
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| :<math>\begin{align}
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| \sum_{n=1}^\infty \frac{1}{n(n+1)} & {} = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) \\
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| & {} = \lim_{N\to\infty} \sum_{n=1}^N \left( \frac{1}{n} - \frac{1}{n+1} \right) \\
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| & {} = \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 \\
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| & {} = \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 = 1.
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| \end{align}</math>
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| ==In general==
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| Let <math>a_n</math> be a sequence of numbers. Then,
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| :<math>\sum_{n=1}^N \left(a_n - a_{n-1}\right) = a_N - a_{0},</math>
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| and, if <math>a_n \rightarrow 0</math>
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| :<math>\sum_{n=1}^\infty \left(a_n - a_{n-1}\right) = - a_{0}.</math>
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| ==A pitfall==
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| Although telescoping can be a useful technique, there are pitfalls to watch out for:
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| :<math>0 = \sum_{n=1}^\infty 0 = \sum_{n=1}^\infty (1-1) = 1 + \sum_{n=1}^\infty (-1 + 1) = 1\,</math>
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| is not correct because this regrouping of terms is invalid unless the individual terms [[Convergent sequence|converge]] to 0; see [[Grandi's series]]. The way to avoid this error is to find the sum of the first ''N'' terms first and ''then'' take the limit as ''N'' approaches infinity: | |
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| :<math>
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| \begin{align}
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| \sum_{n=1}^N \frac{1}{n(n+1)} & {} = \sum_{n=1}^N \left( \frac{1}{n} - \frac{1}{n+1} \right) \\
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| & {} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} -\frac{1}{N+1}\right) \\
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| & {} = 1 + \left(- \frac{1}{2} + \frac{1}{2}\right)
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| + \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots
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| + \left(-\frac{1}{N} + \frac{1}{N}\right) - \frac{1}{N+1} \\
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| & {} = 1 - \frac{1}{N+1}\to 1\ \mathrm{as}\ N\to\infty.
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| \end{align}
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| </math>
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| ==More examples==
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| * Many [[trigonometric function]]s also admit representation as a difference, which allows telescopic cancelling between the consecutive terms.
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| ::<math>
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| \begin{align}
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| \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) \\
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| & {} =\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) \\
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| & {} =\frac{1}{2} \csc\left(\frac{1}{2}\right) \left(\cos\left(\frac{1}{2}\right) -\cos\left(\frac{2N+1}{2}\right)\right).
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| \end{align}
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| </math>
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| * Some sums of the form
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| ::<math>\sum_{n=1}^N {f(n) \over g(n)},</math>
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| :where ''f'' and ''g'' are [[polynomial function]]s whose quotient may be broken up into [[partial fraction]]s, will fail to admit [[summation]] by this method. In particular, we have
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| ::<math>
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| \begin{align}
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| \sum^\infty_{n=0}\frac{2n+3}{(n+1)(n+2)} & {} =\sum^\infty_{n=0}\left(\frac{1}{n+1}+\frac{1}{n+2}\right) \\
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| & {} = \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 \\
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| & {} \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 \\
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| & {} =\infty.
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| \end{align}
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| </math>
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| :The problem is that the terms do not cancel.
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| * Let ''k'' be a positive integer. Then
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| ::<math>\sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k} </math> | |
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| :where ''H''<sub>''k''</sub> is the ''k''th [[harmonic number]]. All of the terms after 1/(''k'' − 1) cancel.
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| == An application in probability theory ==
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| 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
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| : <math> \Pr(X_t = x) = \frac{(\lambda t)^x e^{-\lambda t}}{x!}, </math> | |
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| 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. The density function we seek is therefore
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| : <math>
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| \begin{align}
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| 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)) \\ \\
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| & {} = \frac{d}{dt}\left( 1 - \sum_{u=0}^{x-1} \Pr(X_t = u)\right)
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| = \frac{d}{dt}\left( 1 - \sum_{u=0}^{x-1} \frac{(\lambda t)^u e^{-\lambda t}}{u!} \right) \\ \\
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| & {} = \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)
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| \end{align}
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| </math>
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| The sum telescopes, leaving
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| : <math> f(t) = \frac{\lambda^x t^{x-1} e^{-\lambda t}}{(x-1)!}. </math>
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| == Other applications ==
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| For other applications, see:
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| * [[Grandi's series]];
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| * [[Proof that the sum of the reciprocals of the primes diverges]], where one of the proofs uses a telescoping sum;
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| * [[Order statistic]], where a telescoping sum occurs in the derivation of a probability density function;
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| * [[Lefschetz fixed-point theorem]], where a telescoping sum arises in [[algebraic topology]];
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| * [[Homology theory]], again in algebraic topology;
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| * [[Eilenberg–Mazur swindle]], where a telescoping sum of knots occurs.
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| == Notes and references ==
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| {{reflist}}
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| {{DEFAULTSORT:Telescoping Series}}
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| [[Category:Mathematical series]]
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His name is Merlin but it is not the most masucline name out that there. Guam has for ages been her home but now she is considering likewise. My husband doesn't like it the way I do but the things i really look foward to is driving but I struggle to search out time for this. Production and planning precisely what she executes. Check out one of the most news website: http://www.quantumpendants.org/
My web site: Quantum Energy Pendant Benefits