Hadamard matrix

In mathematics, a telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation. Such a technique is also known as the method of differences.

For example, the series

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

simplifies as

$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}& =\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})\\ & =\underset{N\to \mathrm{\infty }}{\lim }\sum _{n=1}^N(\frac{1}{n}-\frac{1}{n+1})\\ & =\underset{N\to \mathrm{\infty }}{\lim }\left[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots +(\frac{1}{N}-\frac{1}{N+1})\right]\\ & =\underset{N\to \mathrm{\infty }}{\lim }\left[1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+\cdots +(-\frac{1}{N}+\frac{1}{N})-\frac{1}{N+1}\right]=1.\end{array}$

In general

Let $a_n$ be a sequence of numbers. Then,

$\sum _{n=1}^N(a_n-a_{n-1})=a_N-a_0,$

and, if $a_n\to 0$

$\sum _{n=1}^{\mathrm{\infty }}(a_n-a_{n-1})=-a_0.$

A pitfall

Although telescoping can be a useful technique, there are pitfalls to watch out for:

$0=\sum _{n=1}^{\mathrm{\infty }}0=\sum _{n=1}^{\mathrm{\infty }}(1-1)=1+\sum _{n=1}^{\mathrm{\infty }}(-1+1)=1$

is not correct because this regrouping of terms is invalid unless the individual terms converge to 0; see Grandi's series. The way to avoid this error is to find the sum of the first N terms first and then take the limit as N approaches infinity:

$\begin{array}{rl}\sum _{n=1}^N\frac{1}{n(n+1)}& =\sum _{n=1}^N(\frac{1}{n}-\frac{1}{n+1})\\ & =(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots +(\frac{1}{N}-\frac{1}{N+1})\\ & =1+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+\cdots +(-\frac{1}{N}+\frac{1}{N})-\frac{1}{N+1}\\ & =1-\frac{1}{N+1}\to 1\mathrm{a}\mathrm{s}N\to \mathrm{\infty }.\end{array}$

More examples

• Many trigonometric functions also admit representation as a difference, which allows telescopic cancelling between the consecutive terms.

$\begin{array}{rl}\sum _{n=1}^N\sin \left(n\right)& =\sum _{n=1}^N\frac{1}{2}\csc \left(\frac{1}{2}\right)(2\sin \left(\frac{1}{2}\right)\sin \left(n\right))\\ & =\frac{1}{2}\csc \left(\frac{1}{2}\right)\sum _{n=1}^N(\cos \left(\frac{2n-1}{2}\right)-\cos \left(\frac{2n+1}{2}\right))\\ & =\frac{1}{2}\csc \left(\frac{1}{2}\right)(\cos \left(\frac{1}{2}\right)-\cos \left(\frac{2N+1}{2}\right)).\end{array}$

• Some sums of the form

$\sum _{n=1}^N\frac{f(n)}{g(n)},$

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, we have

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{2n+3}{(n+1)(n+2)}& =\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{n+1}+\frac{1}{n+2})\\ & =(\frac{1}{1}+\frac{1}{2})+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{3}+\frac{1}{4})+\cdots \\ & \cdots +(\frac{1}{n-1}+\frac{1}{n})+(\frac{1}{n}+\frac{1}{n+1})+(\frac{1}{n+1}+\frac{1}{n+2})+\cdots \\ & =\mathrm{\infty }.\end{array}$

The problem is that the terms do not cancel.

• Let k be a positive integer. Then

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+k)}=\frac{H_k}{k}$

where H_k is the kth harmonic number. All of the terms after 1/(k − 1) cancel.

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

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

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. The density function we seek is therefore

$\begin{array}{rl}f(t)& =\frac{d}{dt}\mathrm{Pr}(T_x\le t)=\frac{d}{dt}\mathrm{Pr}(X_t\ge x)=\frac{d}{dt}(1-\mathrm{Pr}(X_t\le x-1))\\ \\ & =\frac{d}{dt}(1-\sum _{u=0}^{x-1}\mathrm{Pr}(X_t=u))=\frac{d}{dt}(1-\sum _{u=0}^{x-1}\frac{(\lambda t{)}^u{e}^{-\lambda t}}{u!})\\ \\ & =\lambda e^{-\lambda t}-e^{-\lambda t}\sum _{u=1}^{x-1}(\frac{{\lambda }^u{t}^{u-1}}{(u-1)!}-\frac{{\lambda }^{u+1}{t}^u}{u!})\end{array}$

The sum telescopes, leaving

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

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;
• 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.

Notes and references

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