telescoping series

{{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 whose general term t_n is of the form t_n=a_{n+1}-a_n, i.e. the difference of two consecutive terms of a sequence (a_n). As a consequence the partial sums of the series only consists of two terms of (a_n) after cancellation.{{cite book |last1=Apostol |first1=Tom |title=Calculus, Volume 1 |date=1967 |publisher=John Wiley & Sons |edition=Second |pages=386–387 |orig-date=1961}}Brian S. Thomson and Andrew M. Bruckner, Elementary Real Analysis, Second Edition, CreateSpace, 2008, page 85

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.{{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}}

Definition

File:Telescoping Series.png, \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 partly cancel each other, leaving only parts of the initial and final terms.{{cite web |last1=Weisstein |first1=Eric W. |title=Telescoping Sum |url=https://mathworld.wolfram.com/TelescopingSum.html |website=MathWorld |publisher=Wolfram |language=en}} Let a_n be the elements of a sequence of numbers. Then

\sum_{n=1}^N \left(a_n - a_{n-1}\right) = a_N - a_0.

If a_n converges to a limit L, the telescoping series gives:

\sum_{n=1}^\infty \left(a_n - a_{n-1}\right) = L-a_0.

Every series is a telescoping series of its own partial sums.{{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}}

Examples

  • The product of a geometric series with initial term a and common ratio r by the factor (1 - r) yields a telescoping sum, which allows for a direct calculation of its limit:{{cite book |last1=Apostol |first1=Tom |title=Calculus, Volume 1 |date=1967 |publisher=John Wiley & Sons |edition=Second |page=388 |orig-date=1961}}(1 - r) \sum^\infty_{n=0} ar^n = \sum^\infty_{n=0} \left(ar^n - ar^{n+1}\right) = a when |r| < 1, so when |r| < 1, \sum^\infty_{n=0} ar^n = \frac{a}{1 - r}.

\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}

  • Let k be a positive integer. Then\sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k} where Hk is the kth harmonic number.
  • Let k and m with k \neq m be positive integers. Then\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!} where ! denotes the factorial operation.

\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} which does not converge as N \rightarrow \infty.

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 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

: \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 {Xt ≥ x} is the same as the event {Txt}, and thus they have the same probability. Intuitively, if something occurs at least x times before time t, we have to wait at most t for the xth occurrence. The density function we seek is therefore

:

\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}

The sum telescopes, leaving

: f(t) = \frac{\lambda^x t^{x-1} e^{-\lambda t}}{(x-1)!}.

For other applications, see:

Related concepts

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.{{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}}{{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}} It is the finite products in which consecutive terms cancel denominator with numerator, leaving only the initial and final terms. Let a_n be a sequence of numbers. Then,

\prod_{n=1}^N \frac{a_{n-1}}{a_n} = \frac{a_0}{a_N}.

If a_n converges to 1, the resulting product gives:

\prod_{n=1}^\infty \frac{a_{n-1}}{a_n} = a_0

For example, the infinite product

\prod_{n=2}^{\infty} \left(1-\frac{1}{n^2} \right)

simplifies as

\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}

References

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