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Infinite Product — Definition, Formula & Examples

An infinite product is the result of multiplying infinitely many factors together, written as ∏n=1∞an\prod_{n=1}^{\infty} a_n. It converges if the sequence of partial products PN=a1⋅a2⋯aNP_N = a_1 \cdot a_2 \cdots a_N approaches a finite, nonzero limit.

Given a sequence {an}n=1∞\{a_n\}_{n=1}^{\infty} of real (or complex) numbers, the infinite product ∏n=1∞an\prod_{n=1}^{\infty} a_n is said to converge if the sequence of partial products PN=∏n=1NanP_N = \prod_{n=1}^{N} a_n converges to a limit P≠0P \neq 0. If PN→0P_N \to 0 or the limit does not exist, the product is said to diverge. When every an>0a_n > 0, it is standard to write an=1+bna_n = 1 + b_n and study convergence through the series ∑ln⁡(1+bn)\sum \ln(1 + b_n).

Key Formula

∏n=1∞an=lim⁡N→∞∏n=1Nan\prod_{n=1}^{\infty} a_n = \lim_{N \to \infty} \prod_{n=1}^{N} a_n
Where:
  • ana_n = The nth factor in the product
  • NN = Upper index of the partial product

How It Works

To check whether ∏(1+bn)\prod (1 + b_n) converges (with bn>−1b_n > -1), take logarithms and examine the series ∑ln⁡(1+bn)\sum \ln(1 + b_n). The infinite product converges if and only if this companion series converges. For small bnb_n, ln⁡(1+bn)≈bn\ln(1 + b_n) \approx b_n, so convergence of ∑bn\sum b_n (when all bn>0b_n > 0) implies convergence of the product. Absolute convergence of the product is defined as convergence of ∑∣ln⁡(1+bn)∣\sum |\ln(1 + b_n)|, which holds whenever ∑∣bn∣\sum |b_n| converges.

Worked Example

Problem: Determine whether the infinite product ∏n=2∞(1−1n2)\prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right) converges, and if so, find its value.
Factor each term: Write each factor as a product of two linear pieces.
1−1n2=n2−1n2=(n−1)(n+1)n⋅n1 - \frac{1}{n^2} = \frac{n^2-1}{n^2} = \frac{(n-1)(n+1)}{n \cdot n}
Write out partial products: The partial product telescopes when you separate the numerator and denominator.
PN=∏n=2N(n−1)(n+1)n2=1⋅322⋅2⋅432⋅3⋅542⋯(N−1)(N+1)N2P_N = \prod_{n=2}^{N} \frac{(n-1)(n+1)}{n^2} = \frac{1 \cdot 3}{2^2} \cdot \frac{2 \cdot 4}{3^2} \cdot \frac{3 \cdot 5}{4^2} \cdots \frac{(N-1)(N+1)}{N^2}
Simplify via telescoping: After cancellation, the partial product reduces to a simple expression.
PN=1⋅(N+1)2⋅N=N+12NP_N = \frac{1 \cdot (N+1)}{2 \cdot N} = \frac{N+1}{2N}
Take the limit: As N→∞N \to \infty, the partial product approaches a finite nonzero value.
lim⁡N→∞N+12N=12\lim_{N \to \infty} \frac{N+1}{2N} = \frac{1}{2}
Answer: The infinite product converges to 12\dfrac{1}{2}.

Why It Matters

Infinite products appear throughout analysis and number theory. Euler's product formula expresses the Riemann zeta function as a product over primes, connecting series convergence to the distribution of prime numbers. In complex analysis, entire functions like sin⁡(πz)\sin(\pi z) are represented via Weierstrass product expansions.

Common Mistakes

Mistake: Concluding the product converges when the partial products tend to zero.
Correction: By convention, an infinite product converges only if the limit of partial products is finite and nonzero. A limit of zero counts as divergence (called 'divergence to zero').