Products of near-one characteristic factors

Put $w_{n,k}=z_{n,k}-1$. For $\mathrm{\mid }w\mathrm{\mid }\le 1\mathrm{/}2$, absolute convergence gives $\mathrm{\mid }{e}^w-1-w\mathrm{\mid }\le {\sum }_{j\ge 2}\mathrm{\mid }w{\mathrm{\mid }}^j\mathrm{/}j!\le \mathrm{\mid }w{\mathrm{\mid }}^2{\sum }_{j\ge 2}(1\mathrm{/}2{)}^{j-2}\mathrm{/}2=\mathrm{\mid }w{\mathrm{\mid }}^2$. Also $\mathrm{\mid }1+w\mathrm{\mid }\le 1+\mathrm{\mid }w\mathrm{\mid }\le e^{\mathrm{\mid }w\mathrm{\mid }}$ and $\mathrm{\mid }{e}^w\mathrm{\mid }\le {\sum }_{j\ge 0}\mathrm{\mid }w{\mathrm{\mid }}^j\mathrm{/}j!=e^{\mathrm{\mid }w\mathrm{\mid }}$. The real series is nonnegative term by term, which proves the middle inequality without a logarithm.

For arbitrary finite lists a,b of length r, subtracting successive mixed products gives ${\prod }_{k=1}^r{a}_k-{\prod }_{k=1}^r{b}_k={\sum }_{j=1}^r(a_j-b_j)({\prod }_{k<j}{a}_k)({\prod }_{k>j}{b}_k)$. The cancellation follows by expanding each difference; for r=0 both products are one and the sum zero. Apply it with $a_k=1+w_{n,k}$ and $b_k=e^{w_{n,k}}$. For all sufficiently large n every modulus of w is at most 1/2. Each pair of partial products is bounded by $\exp ({\sum }_{k\mathrm{\ne }j}\mathrm{\mid }{w}_{n,k}\mathrm{\mid })\le e^M$, so the difference of the row products has modulus at most $e^M{\sum }_j\mathrm{\mid }{w}_{n,j}{\mathrm{\mid }}^2$.

The bound tends to zero by hypothesis. Repeated exponential addition identifies the comparison product with $\exp ({\sum }_k{w}_{n,k})$, including empty and one-factor rows. If M=0 then all w vanish and equality is exact in every row. All products and telescoping sums are finite; no branch of logarithm or choice principle is used.