Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors

Let $(a_k)$ be a sequence of reals. Its partial products are

$\Pi_n \;:=\; \prod_{k<n} a_k \qquad (n \in \mathbb{N}),$

the finite products of Finite sums and finite products, by recursion, so that $\Pi_0 = 1$, the empty product, and $\Pi_{n+1} = \Pi_n\, a_n$. For $N \in \mathbb{N}$ the $N$-th tail products are $T^{(N)}_n := \prod_{j<n} a_{N+j}$, again a sequence in $n$.

Convergence. The infinite product $\prod a_k$ converges when there exists $N \in \mathbb{N}$ such that

1. $a_k \ne 0$ for every $k \ge N$, and
2. the sequence $(T^{(N)}_n)_n$ of $N$-th tail products converges (Limits and Cauchy sequences of reals) to a limit $\ell \ne 0$.

Its value is then

$\prod_{k=0}^{\infty} a_k \;:=\; \Bigl(\prod_{k<N} a_k\Bigr)\cdot \ell .$

If no such $N$ exists, the product diverges.

The value does not depend on $N$, and that is a proof obligation, discharged here. First, if $N$ is such an index then so is every $N' \ge N$: condition 1 is inherited, and splitting the finite product (Laws of finite sums and finite products) gives, for $n \ge N' - N$,

$T^{(N)}_{n} \;=\; \Bigl(\prod_{k=N}^{N'-1} a_k\Bigr)\, T^{(N')}_{\,n - (N'-N)} ,$

where the bracketed factor is a product of finitely many nonzero reals and so is itself nonzero. Hence $(T^{(N')}_m)_m$ converges, to $\ell' = \ell / \prod_{k=N}^{N'-1}a_k$ by the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients), and $\ell' \ne 0$ because $\ell \ne 0$. Second, the two candidate values agree:

$\Bigl(\prod_{k<N'}a_k\Bigr)\ell' = \Bigl(\prod_{k<N}a_k\Bigr)\Bigl(\prod_{k=N}^{N'-1}a_k\Bigr)\frac{\ell}{\prod_{k=N}^{N'-1}a_k} = \Bigl(\prod_{k<N}a_k\Bigr)\ell ,$

again by splitting. Finally, any two admissible indices $N_1, N_2$ are both at most $\max\{N_1,N_2\}$, which is therefore admissible and gives the same value as each. Since a convergent sequence has exactly one limit (A sequence has at most one limit), the displayed value is a single well-determined real number.

Why a zero limit is excluded. The definition demands $\ell \ne 0$, not merely that the tail products converge. Both parts of the definition are doing work, and against different naive alternatives. Against the naive "$\Pi_n$ converges", with no tail clause at all: every sequence with a single zero factor has all its partial products equal to $0$ from that index on, hence convergent to $0$, so "the product converges" would say nothing whatever about the factors — which is what condition 1, the restriction to a tail of nonzero factors, repairs. Against the naive "some tail of the partial products converges", which keeps condition 1 and drops only $\ell \ne 0$, condition 1 no longer helps, and a product like $\prod_{j \ge 0}\bigl(1 - 1/(j+2)\bigr)$, all of whose factors are nonzero, has partial products $1/(n+1)$ tending to $0$; calling that convergent would make the value $0$ without any factor being $0$, and would destroy the analogy with series in which a convergent product may be divided by. That product is worked out on the companion examples page.

• The value is $0$ exactly when some factor is $0$. With $N$ as in the definition, the value is $\bigl(\prod_{k<N}a_k\bigr)\ell$ with $\ell \ne 0$, and a finite product vanishes exactly when one of its factors does — a field has no zero divisors (A field has no zero divisors: $ab = 0 \Rightarrow a = 0$ or $b = 0$), so an induction on the recursion $\prod_{k<n+1}a_k = \bigl(\prod_{k<n}a_k\bigr)a_n$ of Finite sums and finite products, by recursion gives both directions. So a convergent product with all factors nonzero has nonzero value, and this is the property that makes convergent products behave like nonzero numbers.

• Finitely many factors may be $0$, or negative, or anything at all. The definition looks only at a tail, exactly as Series, partial sums, convergence and the sum, divergence, and the tail series does for series through its tail clause; conditions 1 and 2 constrain no initial segment.

• Notation. $\prod_{k \ge m} a_k$ denotes the product of the family from $m$, that is the product of the sequence $j \mapsto a_{j+m}$, by the same convention Series, partial sums, convergence and the sum, divergence, and the tail series uses for series; the two readings agree at $m = 0$.

• Nothing here presumes a logarithm. The classical criteria for infinite products are usually derived by taking logarithms; the logarithm is not available at this point in the reading order, and For $p_k \ge 0$ the product $\prod (1 + p_k)$ converges iff $\sum p_k$ converges, with $1 + \sum_{k<n} p_k \le \prod_{k<n}(1+p_k) \le 1/\bigl(1 - \sum_{k<n} p_k\bigr)$ when $\sum_{k<n} p_k < 1$; for $0 \le p_k < 1$ the product $\prod (1 - p_k)$ converges iff $\sum p_k$ converges and its partial products tend to $0$ otherwise; and $\sum |p_k|$ convergent implies $\prod (1+p_k)$ convergent is proved from elementary inequalities instead.