Telescoping sums are finite sums in which pairs of consecutive terms cancel each other, leaving only the initial and final terms.^[4]

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

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

• 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
  $${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(n+k)(n+k+1)\dots (n+m-1)(n+m)}}={\frac {1}{m-k}}\cdot {\frac {k!}{m!}}}$$

   

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

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

Telescoping productEdit

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.^[5][6]

For example, the infinite product^[5]

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

simplifies as

${\displaystyle {\begin{aligned}\prod _{n=2}^{\infty }\left(1-{\frac {1}{n^{2}}}\right)&=\prod _{n=2}^{\infty }{\frac {(n-1)(n+1)}{n^{2}}}\\&=\lim _{N\to \infty }\prod _{n=2}^{N}{\frac {n-1}{n}}\times \prod _{n=2}^{N}{\frac {n+1}{n}}\\&=\lim _{N\to \infty }\left\lbrack {{\frac {1}{2}}\times {\frac {2}{3}}\times {\frac {3}{4}}\times \cdots \times {\frac {N-1}{N}}}\right\rbrack \times \left\lbrack {{\frac {3}{2}}\times {\frac {4}{3}}\times {\frac {5}{4}}\times \cdots \times {\frac {N}{N-1}}\times {\frac {N+1}{N}}}\right\rbrack \\&=\lim _{N\to \infty }\left\lbrack {\frac {1}{2}}\right\rbrack \times \left\lbrack {\frac {N+1}{N}}\right\rbrack \\&={\frac {1}{2}}\times \lim _{N\to \infty }\left\lbrack {\frac {N+1}{N}}\right\rbrack \\&={\frac {1}{2}}\times \lim _{N\to \infty }\left\lbrack {\frac {N}{N}}+{\frac {1}{N}}\right\rbrack \\&={\frac {1}{2}}.\end{aligned}}}$ 

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.