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

${\mathrm{\Pi }}_n:={\prod }_{k<n}{a}_k(n\in \mathbb{N}),$

the finite products of Finite sums and finite products, by recursion, so that ${\mathrm{\Pi }}_0=1$, the empty product, and ${\mathrm{\Pi }}_{n+1}={\mathrm{\Pi }}_n{a}_n$. 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. $a_k\mathrm{\ne }0$ 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 $\mathrm{\ell }\mathrm{\ne }0$.

Its value is then

${\prod }_{k=0}^{\mathrm{\infty }}{a}_k:=\left({\prod }_{k<N}{a}_k\right)\cdot \mathrm{\ell }\mathrm{.}$

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 $N^{\mathrm{\prime }}\ge N$: condition 1 is inherited, and splitting the finite product (Laws of finite sums and finite products) gives, for $n\ge N^{\mathrm{\prime }}-N$,

$T_n^{(N)}=\left({\prod }_{k=N}^{N^{\mathrm{\prime }}-1}{a}_k\right)T_{n-(N^{\mathrm{\prime }}-N)}^{(N^{\mathrm{\prime }})},$

where the bracketed factor is a product of finitely many nonzero reals and so is itself nonzero. Hence $(T_m^{(N^{\mathrm{\prime }})}{)}_m$ converges, to ${\mathrm{\ell }}^{\mathrm{\prime }}=\mathrm{\ell }\mathrm{/}{\prod }_{k=N}^{N^{\mathrm{\prime }}-1}{a}_k$ by the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients), and ${\mathrm{\ell }}^{\mathrm{\prime }}\mathrm{\ne }0$ because $\mathrm{\ell }\mathrm{\ne }0$. Second, the two candidate values agree:

$\left({\prod }_{k<N^{\mathrm{\prime }}}{a}_k\right){\mathrm{\ell }}^{\mathrm{\prime }}=\left({\prod }_{k<N}{a}_k\right)\left({\prod }_{k=N}^{N^{\mathrm{\prime }}-1}{a}_k\right)\frac{\mathrm{\ell }}{{\prod }_{k=N}^{N^{\mathrm{\prime }}-1}{a}_k}=\left({\prod }_{k<N}{a}_k\right)\mathrm{\ell },$

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 $\mathrm{\ell }\mathrm{\ne }0$, not merely that the tail products converge. Both parts of the definition are doing work, and against different naive alternatives. Against the naive "${\mathrm{\Pi }}_n$ 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 $\mathrm{\ell }\mathrm{\ne }0$, condition 1 no longer helps, and a product like ${\prod }_{j\ge 0}(1-1\mathrm{/}(j+2))$, all of whose factors are nonzero, has partial products $1\mathrm{/}(n+1)$ 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 $\left({\prod }_{k<N}{a}_k\right)\mathrm{\ell }$ with $\mathrm{\ell }\mathrm{\ne }0$, 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 ${\prod }_{k<n+1}{a}_k=\left({\prod }_{k<n}{a}_k\right)a_n$ 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 $1+{\sum }_{k<n}{p}_k\le {\prod }_{k<n}(1+p_k)\le 1\mathrm{/}(1-{\sum }_{k<n}{p}_k)$ 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 $\sum \mathrm{\mid }{p}_k\mathrm{\mid }$ convergent implies $\prod (1+p_k)$ convergent is proved from elementary inequalities instead.