telescoping series

File:Telescoping Series.png, $\backslash sum$, 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

$$\backslash sum\_\{n=1\}^N\; \backslash left(a\_n\; -\; a\_\{n-1\}\backslash right)\; =\; a\_N\; -\; a\_0.$$

If $a\_n$ converges to a limit $L$, the telescoping series gives:

$$\backslash sum\_\{n=1\}^\backslash infty\; \backslash left(a\_n\; -\; a\_\{n-1\}\backslash right)\; =\; L-a\_0.$$

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}}

• 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}}$$(1\; -\; r)\; \backslash sum^\backslash infty\_\{n=0\}\; ar^n\; =\; \backslash sum^\backslash infty\_\{n=0\}\; \backslash left(ar^n\; -\; ar^\{n+1\}\backslash right)\; =\; a$$when so when $$\backslash sum^\backslash infty\_\{n=0\}\; ar^n\; =\; \backslash frac\{a\}\{1\; -\; r\}.$$

• The series$$\backslash sum\_\{n=1\}^\backslash infty\backslash frac\{1\}\{n(n+1)\}$$is the series of reciprocals of pronic numbers, and it is recognizable as a telescoping series once rewritten in partial fraction form $\backslash begin\{align\}$

\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}

• Let k be a positive integer. Then$$\backslash sum^\backslash infty\_\{n=1\}\; \{\backslash frac\{1\}\{n(n+k)\}\}\; =\; \backslash frac\{H\_k\}\{k\}$$ where H_k is the kth harmonic number.

• Let k and m with k $\backslash neq$ m be positive integers. Then$$\backslash sum^\backslash infty\_\{n=1\}\; \{\backslash frac\{1\}\{(n+k)(n+k+1)\backslash dots(n+m-1)(n+m)\}\}\; =\; \backslash frac\{1\}\{m-k\}\; \backslash cdot\; \backslash frac\{k!\}\{m!\}$$ where $!$ 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,$$\backslash begin\{align\}$$

\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} which does not converge as $N\; \backslash rightarrow\; \backslash infty.$

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 xth "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

: $\backslash Pr(X\_t\; =\; x)\; =\; \backslash frac\{(\backslash lambda\; t)^x\; e^\{-\backslash lambda\; t\}\}\{x!\},$

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 $x$ 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

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

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.

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,

$$\backslash prod\_\{n=1\}^N\; \backslash frac\{a\_\{n-1\}\}\{a\_n\}\; =\; \backslash frac\{a\_0\}\{a\_N\}.$$

If $a\_n$ converges to 1, the resulting product gives:

$$\backslash prod\_\{n=1\}^\backslash infty\; \backslash frac\{a\_\{n-1\}\}\{a\_n\}\; =\; a\_0$$

For example, the infinite product

$$\backslash prod\_\{n=2\}^\{\backslash infty\}\; \backslash left(1-\backslash frac\{1\}\{n^2\}\; \backslash right)$$

simplifies as

$$\backslash begin\{align\}$$

\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}

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