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

As a consequence the partial sums only consists of two terms of ${\displaystyle (a_{n})}$ after cancellation.^[2][3] 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

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

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.^[4]

In general

Telescoping sums are finite sums in which pairs of consecutive terms cancel each other, leaving only the initial and final terms.^[5]

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 canceling 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 H_k is the kth harmonic number. All of the terms after 1/(k − 1) cancel.
• Let k,m with k ${\displaystyle \neq }$ m be positive integers. Then
  $${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(n+k)(n+k+1)\dots (n+m-1)(n+m)))={\frac {1}{m-k))\cdot {\frac {k!}{m!))}$$

  

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 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

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

where λ is the average number of occurrences in any time interval of length 1. Observe that the event {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 ${\displaystyle x}$ 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.^[6][7]

For example, the infinite product^[6]

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

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:

• Grandi's series;
• 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.