How To Find The Sum Of A Telescoping Series

Serial whose partial sums eventually only have a fixed number of terms later on cancellation

In mathematics, a telescoping series is a series whose full general term ${\displaystyle t_{due north}}$ can be written as ${\displaystyle t_{n}=a_{n}-a_{n+one}}$ , i.e. the difference of ii consecutive terms of a sequence ${\displaystyle (a_{n})}$ .^{[ citation needed ]}

As a consequence the partial sums just consists of two terms of ${\displaystyle (a_{due north})}$ after cancellation.^{[i] [ii]} The counterfoil technique, with part of each term cancelling with part of the next term, is known as the method of differences.

For example, the series

${\displaystyle \sum _{north=1}^{\infty }{\frac {1}{n(n+one)}}}$

(the series of reciprocals of pronic numbers) simplifies as

${\displaystyle {\begin{aligned}\sum _{n=i}^{\infty }{\frac {1}{north(n+1)}}&{}=\sum _{north=1}^{\infty }\left({\frac {1}{northward}}-{\frac {one}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\sum _{n=ane}^{Northward}\left({\frac {1}{north}}-{\frac {1}{n+i}}\right)\\{}&{}=\lim _{N\to \infty }\left\lbrack {\left(i-{\frac {i}{2}}\right)+\left({\frac {one}{2}}-{\frac {ane}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {i}{Due north+1}}\right)}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1+\left(-{\frac {ane}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {one}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {ane}{Northward}}+{\frac {one}{Due north}}\right)-{\frac {1}{Northward+one}}}\right\rbrack \\{}&{}=\lim _{Northward\to \infty }\left\lbrack {ane-{\frac {1}{N+one}}}\right\rbrack =1.\end{aligned}}}$

An early on argument of the formula for the sum or fractional sums of a telescoping serial tin exist found in a 1644 piece of work by Evangelista Torricelli, De dimensione parabolae.^[three]

In general [edit]

A telescoping series of powers. Annotation in the summation sign, ${\textstyle \sum }$ , the index due north goes from 1 to m. In that location is no relationship betwixt n and m beyond the fact that both are natural numbers.

Telescoping sums are finite sums in which pairs of sequent terms abolish each other, leaving only the initial and last terms.^[4]

Let ${\displaystyle a_{n}}$ be a sequence of numbers. Then,

${\displaystyle \sum _{n=1}^{North}\left(a_{north}-a_{northward-one}\right)=a_{N}-a_{0}}$

If ${\displaystyle a_{n}\rightarrow 0}$

${\displaystyle \sum _{n=one}^{\infty }\left(a_{n}-a_{n-ane}\right)=-a_{0}}$

Telescoping products are finite products in which sequent terms cancel denominator with numerator, leaving just the initial and final terms.

Permit ${\displaystyle a_{northward}}$ be a sequence of numbers. Then,

${\displaystyle \prod _{due north=1}^{N}{\frac {a_{north-one}}{a_{n}}}={\frac {a_{0}}{a_{Due north}}}}$

If ${\displaystyle a_{n}\rightarrow 1}$

${\displaystyle \prod _{n=one}^{\infty }{\frac {a_{north-1}}{a_{n}}}=a_{0}}$

More examples [edit]

• Many trigonometric functions too admit representation as a divergence, which allows scope canceling betwixt the consecutive terms.

  $${\displaystyle {\begin{aligned}\sum _{n=1}^{N}\sin \left(northward\right)&{}=\sum _{n=1}^{N}{\frac {one}{2}}\csc \left({\frac {1}{2}}\correct)\left(ii\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\correct)\\&{}={\frac {1}{ii}}\csc \left({\frac {one}{ii}}\right)\sum _{north=1}^{N}\left(\cos \left({\frac {2n-1}{2}}\correct)-\cos \left({\frac {2n+ane}{2}}\right)\correct)\\&{}={\frac {1}{2}}\csc \left({\frac {1}{two}}\right)\left(\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+one}{2}}\correct)\right).\end{aligned}}}$$

  

• Some sums of the form

  $${\displaystyle \sum _{n=1}^{N}{f(n) \over g(northward)}}$$

  

  where f and g are polynomial functions whose quotient may exist broken up into fractional fractions, will fail to admit summation by this method. In detail, one has

  $${\displaystyle {\brainstorm{aligned}\sum _{n=0}^{\infty }{\frac {2n+3}{(n+i)(n+ii)}}={}&\sum _{n=0}^{\infty }\left({\frac {1}{northward+1}}+{\frac {1}{n+2}}\right)\\={}&\left({\frac {ane}{one}}+{\frac {one}{2}}\right)+\left({\frac {1}{2}}+{\frac {1}{3}}\correct)+\left({\frac {1}{iii}}+{\frac {1}{4}}\right)+\cdots \\&{}\cdots +\left({\frac {ane}{due north-1}}+{\frac {ane}{n}}\correct)+\left({\frac {1}{n}}+{\frac {1}{n+1}}\correct)+\left({\frac {1}{n+i}}+{\frac {1}{north+two}}\right)+\cdots \\={}&\infty .\finish{aligned}}}$$

  

  The problem is that the terms do not abolish.
• Let k be a positive integer. Then

  $${\displaystyle \sum _{n=1}^{\infty }{\frac {ane}{n(n+k)}}={\frac {H_{grand}}{k}}}$$

  

  where H _k is the kth harmonic number. All of the terms after 1/(yard − one) cancel.
• Allow k,m with k ${\displaystyle \neq }$ 1000 be positive integers. Then

  $${\displaystyle \sum _{n=ane}^{\infty }{\frac {1}{(n+k)(n+1000+1)\dots (n+thousand-1)(due north+one thousand)}}={\frac {one}{m-k}}\cdot {\frac {g!}{one thousand!}}}$$

  

An application in probability theory [edit]

In probability theory, a Poisson process is a stochastic process of which the simplest instance involves "occurrences" at random times, the waiting time until the adjacent occurrence having a memoryless exponential distribution, and the number of "occurrences" in whatsoever time interval having a Poisson distribution whose expected value is proportional to the length of the time interval. Let X _t exist the number of "occurrences" before time t, and permit T _x be the waiting time until the tenth "occurrence". We seek the probability density office of the random variable T _x . Nosotros use the probability mass function for the Poisson distribution, which tells us that

${\displaystyle \Pr(X_{t}=x)={\frac {(\lambda t)^{x}east^{-\lambda t}}{10!}},}$

where λ is the average number of occurrences in any time interval of length 1. Observe that the consequence {X _t ≥ x} is the same as the event {T _x ≤ t}, and thus they have the same probability. Intuitively, if something occurs at to the lowest degree ${\displaystyle x}$ times before time ${\displaystyle t}$ , we have to wait at most ${\displaystyle t}$ for the ${\displaystyle xth}$ occurrence. The density function we seek is therefore

${\displaystyle {\begin{aligned}f(t)&{}={\frac {d}{dt}}\Pr(T_{x}\leq t)={\frac {d}{dt}}\Pr(X_{t}\geq x)={\frac {d}{dt}}(1-\Pr(X_{t}\leq x-one))\\\\&{}={\frac {d}{dt}}\left(1-\sum _{u=0}^{x-1}\Pr(X_{t}=u)\right)={\frac {d}{dt}}\left(ane-\sum _{u=0}^{x-ane}{\frac {(\lambda t)^{u}east^{-\lambda t}}{u!}}\right)\\\\&{}=\lambda e^{-\lambda t}-e^{-\lambda t}\sum _{u=1}^{x-one}\left({\frac {\lambda ^{u}t^{u-one}}{(u-1)!}}-{\frac {\lambda ^{u+one}t^{u}}{u!}}\right)\end{aligned}}}$

The sum telescopes, leaving

${\displaystyle f(t)={\frac {\lambda ^{x}t^{x-1}eastward^{-\lambda t}}{(10-one)!}}.}$

Similar concepts [edit]

Telescoping product [edit]

A telescoping production is a finite product (or the partial product of an space production) that tin be cancelled by method of quotients to be somewhen only a finite number of factors.^{[5] [6]}

For example, the space product^[five]

${\displaystyle \prod _{n=2}^{\infty }\left(i-{\frac {i}{n^{2}}}\right)}$

simplifies every bit

${\displaystyle {\brainstorm{aligned}\prod _{n=2}^{\infty }\left(1-{\frac {1}{n^{ii}}}\right)&=\prod _{due north=2}^{\infty }{\frac {(north-1)(n+i)}{n^{2}}}\\&=\lim _{Due north\to \infty }\prod _{n=ii}^{N}{\frac {n-1}{north}}\times \prod _{n=two}^{N}{\frac {due north+one}{n}}\\&=\lim _{North\to \infty }\left\lbrack {{\frac {1}{2}}\times {\frac {two}{3}}\times {\frac {three}{iv}}\times \cdots \times {\frac {N-ane}{N}}}\right\rbrack \times \left\lbrack {{\frac {3}{two}}\times {\frac {four}{iii}}\times {\frac {5}{4}}\times \cdots \times {\frac {N}{N-one}}\times {\frac {N+1}{North}}}\right\rbrack \\&=\lim _{N\to \infty }\left\lbrack {\frac {i}{two}}\correct\rbrack \times \left\lbrack {\frac {N+one}{N}}\right\rbrack \\&={\frac {one}{2}}\times \lim _{N\to \infty }\left\lbrack {\frac {Northward+1}{N}}\right\rbrack \\&={\frac {1}{two}}\times \lim _{North\to \infty }\left\lbrack {\frac {N}{Due north}}+{\frac {i}{N}}\right\rbrack \\&={\frac {one}{2}}.\end{aligned}}}$

Other applications [edit]

For other applications, see:

• Grandi'due south series;
• Proof that the sum of the reciprocals of the primes diverges, where i of the proofs uses a telescoping sum;
• Central 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.

References [edit]

1. ^ Tom M. Apostol, Calculus, Volume i, Blaisdell Publishing Company, 1962, pages 422–3
2. ^ Brian S. Thomson and Andrew M. Bruckner, Simple Real Analysis, 2d Edition, CreateSpace, 2008, page 85
3. ^ Weil, André (1989). "Prehistory of the zeta-function". In Aubert, Karl Egil; Bombieri, Enrico; Goldfeld, Dorian (eds.). Number Theory, Trace Formulas and Detached Groups: Symposium in Honor of Atle Selberg, Oslo, Norway, July 14–21, 1987. Boston, Massachusetts: Academic Press. pp. 1–9. doi:x.1016/B978-0-12-067570-8.50009-3. MR 0993308.
4. ^ Weisstein, Eric W. "Telescoping Sum". MathWorld. Wolfram.
5. ^ ^{a b} "Telescoping Series - Product". Bright Math & Science Wiki. Brilliant.org. Retrieved nine February 2020.
6. ^ Bogomolny, Alexander. "Telescoping Sums, Serial and Products". Cutting the Knot . Retrieved 9 February 2020.

Source: https://en.wikipedia.org/wiki/Telescoping_series

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