Telescoping series

Short description: Series whose partial sums eventually only have a fixed number of terms after cancellation

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

As a consequence the partial sums only consists of two terms of [math]\displaystyle{ (a_n) }[/math] after cancellation. 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

[math]\displaystyle{ \sum_{n=1}^\infty\frac{1}{n(n+1)} }[/math]

(the series of reciprocals of pronic numbers) simplifies as

[math]\displaystyle{ \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} }[/math]

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.

In general

A telescoping series of powers. Note in the summation sign, [math]\displaystyle{ \sum }[/math], 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 cancel each other, leaving only the initial and final terms.

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers. Then,

[math]\displaystyle{ \sum_{n=1}^N \left(a_n - a_{n-1}\right) = a_N - a_0 }[/math]

If [math]\displaystyle{ a_n \rightarrow 0 }[/math]

[math]\displaystyle{ \sum_{n=1}^\infty \left(a_n - a_{n-1}\right) = - a_0 }[/math]

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

Let [math]\displaystyle{ a_n }[/math] be a sequence of numbers. Then,

[math]\displaystyle{ \prod_{n=1}^N \frac{a_{n-1}}{a_n} = \frac{a_0}{a_N} }[/math]

If [math]\displaystyle{ a_n \rightarrow 1 }[/math]

[math]\displaystyle{ \prod_{n=1}^\infty \frac{a_{n-1}}{a_n} = a_0 }[/math]

More examples

• Many trigonometric functions also admit representation as a difference, which allows telescopic canceling between the consecutive terms. [math]\displaystyle{ \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} }[/math]
• Some sums of the form [math]\displaystyle{ \sum_{n=1}^N {f(n) \over g(n)} }[/math] 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 [math]\displaystyle{ \begin{align} \sum^\infty_{n=0}\frac{2n+3}{(n+1)(n+2)} = {} & \sum^\infty_{n=0}\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{align} }[/math] The problem is that the terms do not cancel.
• Let k be a positive integer. Then [math]\displaystyle{ \sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k} }[/math] where H_k is the kth harmonic number. All of the terms after 1/(k − 1) cancel.
• Let k,m with k [math]\displaystyle{ \neq }[/math] m be positive integers. Then [math]\displaystyle{ \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!} }[/math]

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

[math]\displaystyle{ \Pr(X_t = x) = \frac{(\lambda t)^x e^{-\lambda t}}{x!}, }[/math]

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 [math]\displaystyle{ x }[/math] times before time [math]\displaystyle{ t }[/math], we have to wait at most [math]\displaystyle{ t }[/math] for the [math]\displaystyle{ xth }[/math] occurrence. The density function we seek is therefore

[math]\displaystyle{ \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} }[/math]

The sum telescopes, leaving

[math]\displaystyle{ f(t) = \frac{\lambda^x t^{x-1} e^{-\lambda t}}{(x-1)!}. }[/math]

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.

For example, the infinite product

[math]\displaystyle{ \prod_{n=2}^{\infty} \left(1-\frac{1}{n^2} \right) }[/math]

simplifies as

[math]\displaystyle{ \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}\times \lim_{N\to\infty} \left\lbrack \frac{N}{N} + \frac{1}{N} \right\rbrack \\ &=\frac{1}{2}. \end{align} }[/math]

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.