In mathematics, a telescoping product is a product of a sequence of factors that simplifies dramatically because most intermediate terms cancel, leaving only the initial and final terms. This phenomenon is closely related to the concept of a telescoping sum, but applies to multiplicative structures instead of additive ones. It is also known as a method of quotients due to its interpretation as repeated division of successive terms.
Definition Let A ⊆ Z {\displaystyle A\subseteq \mathbb {Z} } and let g : A → R {\displaystyle g:A\to \mathbb {R} } be a function such that g ( n ) ≠ 0 {\displaystyle g(n)\neq 0} for all relevant n {\displaystyle n} . A sequence f ( n ) {\displaystyle f(n)} is said to define a telescoping product if it can be written in the form
f ( n ) = g ( n + 1 ) g ( n ) . {\displaystyle f(n)={\frac {g(n+1)}{g(n)}}.}
Then the finite product
∏ k = m n f ( k ) {\displaystyle \prod _{k=m}^{n}f(k)}
collapses as intermediate terms cancel. This is why the telescoping product is often referred to as the method of quotients: each factor represents a quotient of successive values of a function.
Fundamental identity If g ( k ) ≠ 0 {\displaystyle g(k)\neq 0} for all k {\displaystyle k} , then:
∏ k = m n g ( k + 1 ) g ( k ) = g ( n + 1 ) g ( m ) . {\displaystyle \prod _{k=m}^{n}{\frac {g(k+1)}{g(k)}}={\frac {g(n+1)}{g(m)}}.}
All intermediate terms cancel pairwise, leaving only the boundary terms.
Interpretation The telescoping product can be viewed as the multiplicative analogue of the finite difference method used in telescoping sums. While sums use differences g ( n + 1 ) − g ( n ) {\displaystyle g(n+1)-g(n)} , products use ratios g ( n + 1 ) / g ( n ) {\displaystyle g(n+1)/g(n)} . In both cases, the structure ensures that intermediate contributions vanish in the final expression.
Examples
Basic example Let
f ( n ) = n + 1 n . {\displaystyle f(n)={\frac {n+1}{n}}.}
Then
∏ k = 1 n k + 1 k = 2 1 ⋅ 3 2 ⋅ 4 3 ⋯ n + 1 n = n + 1. {\displaystyle \prod _{k=1}^{n}{\frac {k+1}{k}}={\frac {2}{1}}\cdot {\frac {3}{2}}\cdot {\frac {4}{3}}\cdots {\frac {n+1}{n}}=n+1.}
All intermediate factors cancel, leaving only the final numerator and initial denominator.
Factorial-related example Let
f ( n ) = n n − 1 ( n ≥ 2 ) . {\displaystyle f(n)={\frac {n}{n-1}}\quad (n\geq 2).}
Then
∏ k = 2 n k k − 1 = n 1 = n . {\displaystyle \prod _{k=2}^{n}{\frac {k}{k-1}}={\frac {n}{1}}=n.}
This is another direct consequence of successive cancellation.
Example using a general sequence For any sequence g ( n ) {\displaystyle g(n)} with nonzero terms:
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