# Telescoping series

In mathematics, a telescoping series is a series whose general term ${\displaystyle t_{n}}$ can be written as ${\displaystyle t_{n}=a_{n}-a_{n+1}}$, i.e. the difference of two consecutive terms of a sequence ${\displaystyle (a_{n})}$.[citation needed]

As a consequence the partial sums only consists of two terms of ${\displaystyle (a_{n})}$ after cancellation.[1][2] The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences.

For example, the series

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

(the series of reciprocals of pronic numbers) simplifies as

{\displaystyle {\begin{aligned}\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{aligned}}}

## In general

A telescoping series of powers

Telescoping sums are finite sums in which pairs of consecutive terms cancel each other, leaving only the initial and final terms.[3]

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

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

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

${\displaystyle \sum _{n=1}^{\infty }\left(a_{n}-a_{n-1}\right)=-a_{0}}$

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

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

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

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

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

## More examples

• Many trigonometric functions also admit representation as a difference, which allows telescopic cancelling between the consecutive terms.
{\displaystyle {\begin{aligned}\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{aligned}}}
• Some sums of the form
${\displaystyle \sum _{n=1}^{N}{f(n) \over g(n)}}$
where f and g are polynomial functions whose quotient may be broken up into partial fractions, will fail to admit summation by this method. In particular, one has
{\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }{\frac {2n+3}{(n+1)(n+2)}}={}&\sum _{n=0}^{\infty }\left({\frac {1}{n+1}}+{\frac {1}{n+2}}\right)\\={}&\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 \\&{}\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 \\={}&\infty .\end{aligned}}}
The problem is that the terms do not cancel.
• Let k be a positive integer. Then
${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+k)}}={\frac {H_{k}}{k}}}$
where Hk is the kth harmonic number. All of the terms after 1/(k − 1) cancel.

## An application in probability theory

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

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

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 ${\displaystyle x}$ 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-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 ^{u}t^{u-1}}{(u-1)!}}-{\frac {\lambda ^{u+1}t^{u}}{u!}}\right)\end{aligned}}}

The sum telescopes, leaving

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

## Similar concepts

### Telescoping product

A telescoping product is a finite product (or the partial product of an infinite product) that can be cancelled by method of quotients to be eventually only a finite number of factors.[4][5]

For example, the infinite product[4]

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

simplifies as

{\displaystyle {\begin{aligned}\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}}\times \lim _{N\to \infty }\left\lbrack {\frac {N}{N}}+{\frac {1}{N}}\right\rbrack \\&={\frac {1}{2}}.\end{aligned}}}

## Other applications

For other applications, see:

## References

1. ^ Tom M. Apostol, Calculus, Volume 1, Blaisdell Publishing Company, 1962, pages 422–3
2. ^ Brian S. Thomson and Andrew M. Bruckner, Elementary Real Analysis, Second Edition, CreateSpace, 2008, page 85
3. ^ Weisstein, Eric W. "Telescoping Sum". MathWorld. Wolfram.
4. ^ a b "Telescoping Series - Product". Brilliant Math & Science Wiki. Brilliant.org. Retrieved 9 February 2020.
5. ^ Bogomolny, Alexander. "Telescoping Sums, Series and Products". Cut the Knot. Retrieved 9 February 2020.