The convergence criteria for infinite products can be used to check whether the product converges. This convergence is then assumed to be non-zero.
Question - why does convergence indicated by a convergence test imply a non-zero limit?
My previous related questions have had replies which say the definition of a convergent infinite product is that the limit of the partial products is finite and non-zero.
This seems arbitrary and not an answer.
Having spent many hours, I can't see how the derivations of convergence criteria encode a non-zero limit.
I will give an example below of such a derivation.
1. Convergence Criterion for Real $a_n>1$
Since the terms in a convergent infinite product tend to 1, it is useful to write the factors as $(1+a_{n})$.
$$P=\prod\left(1+a_{n}\right)$$
We can turn a product into a sum by taking the logarithm.
$$\ln(P)=\ln\prod\left(1+a_{n}\right)=\sum\ln\left(1+a_{n}\right)$$
Using $1+x\leq e^{x}$ we arrive at a nice inequality.
$$\ln(P)\leq\sum a_{n}$$
This tells us that if the sum is bounded, the product is bounded too. If the terms $a_{n}$ are always positive, then the sum can only grow monotonically (without oscillation), so the boundedness is convergence. This is a useful result but we can stregthen it.
Expanding out the product $\prod(1+a_{n})$ gives us a sum which includes the terms 1, all the individual $a_{n}$, and also the combinations of different $a_{n}$ multiplied together. This gives us another inequality.
$$1+\sum a_{n}\leq\prod\left(1+a_{n}\right)=P$$
This tell us that if the product converges, so does the sum. The two results together give us our first convergence criterion.
$$\boxed{\sum a_{n}\text{ converges }\Leftrightarrow\prod\left(1+a_{n}\right)\text{ converges, for }a_{n}>0}$$
Note: Nowhere in this derivation is it required that $P\neq 0$
2. Convergence Criterion for Complex $a_n\neq-1$
We start by saying that since $|a_n|>0$, the previous criteria immediately give us:
$$\boxed{\sum|a_{n}|\text{ converges }\Leftrightarrow\prod(1+|a_{n}|)\text{ converges}}$$
which is valid for complex $a_n$.
We are interested in products $\prod(1+a_{n})$ with complex $a_{n}$, not just $\prod(1+|a_{n}|)$. The key to making this leap is the well known fact that if a series converges absolutely, it also converges.
$$\boxed{\sum|a_{n}|\text{ converges }\implies\sum a_{n}\text{ converges}}$$
Let's start with two partial products.
$$p_{N} =\prod^{N}(1+a_{n})$$
$$q_{N} =\prod^{N}(1+|a_{n}|)$$
We need to assert that $a_{n}\neq-1$ to ensure no zero-valued factors $(1+a_{n})$.
For $N>M\geq1$, we can compare $|p_{N}-p_{M}|$ with $|q_{N}-q_{M}|$ with a some simple algebra.
$$\begin{align}\left|p_{N}-p_{M}\right| &=|p_{M}|\cdot\left|\frac{p_{N}}{p_{M}}-1\right| \\ &=|p_{M}|\cdot\left|\prod_{M+1}^{N}(1+a_{n})-1\right| \\ &\leq|q_{M}|\cdot\left|\prod_{M+1}^{N}(1+|a_{n}|)-1\right| \\ &=|q_{M}|\cdot\left|\frac{q_{N}}{q_{M}}-1\right| \\ &=\left|q_{N}-q_{M}\right|\end{align}$$
If $|q_{N}-q_{M}|<\epsilon$, where \epsilon is as small as we want, then $|p_{N}-p_{M}|<\epsilon$ too. This the Cauchy criterion for convergence, and it tells us that if $q_{N}$ converges, so does $p_{N}$.
And we can say $p_N$ converges if $\sum a_n$ converges, which it does if $\sum |a_n|$ converges.
We finally have our third convergence criterion.
$$\boxed{\sum|a_{n}|\text{ converges }\implies\prod(1+a_{n})\text{ converges, for }a_{n}\neq-1}$$
Note: Again, there is nothing in this derivation that requires $p_N$ to converge to a non-zero value.
Example of Use of Criteria to Assert Non-Zero Limit
The Riemann Zeta function can be written as an Euler product:
$$\zeta(s)=\sum\frac{1}{n^{s}}=\prod\left(1-\frac{1}{p^{s}}\right)^{-1}$$
The above convergence criteria can be used with $|a_n| = |-1/p^s$.
$$\sum\left|-\frac{1}{p^{s}}\right|=\sum\frac{1}{p^{\sigma}}\leq\sum\frac{1}{n^{\sigma}}$$
The reason $\sum1/p^{\sigma}\leq\sum1/n^{\sigma}$ is because there are fewer primes $p$ than integers $n$.
Because $\sum1/n^{\sigma}$ converges for $\sigma>1$, so does $\sum|-1/p^{s}|$. This means $1/\zeta(s)$ converges to a non-zero value, and therefore so does $\zeta(s)$.
We can now say the Riemann Zeta function has no zeros in the domain $\sigma>1$.
Note - "converges to a non-zero value" is the contentious statement.
This question is NOT the same as this one Proof that infinite products, if they converge, converge to a non-zero value.
Whenever $\sum |a_n| < \infty, |a_n| < 1/2$, it follows that $1 \ge P_0=\Pi (1 - |a_n|) >0 $ because one can bound $P_0 \ge e^{\sum {-2|a_n|}}>0$; similarly $\infty > P_1=\Pi (1 + |a_n|) \ge 1 > 0$ because $P_1 \le e^{\sum {|a_n|}}< \infty$
This immediately implies that if $\sum |a_n| < \infty, a_n \ne \pm 1$ we have that $Q=\Pi (1 \pm a_n) \ne 0$ (arbitrary choice of signs for each $n$) because if $Q_1$ is the product of the factors of $Q$ where we ignore the finitely many $|a_n| \ge 1/2$ we have that $P_1 \ge |Q_1|\ge P_0$ since $1-|a_n| \le |1 \pm a_n| \le 1+|a_n|$, while convergence of partial products of $Q_1$ follows from the usual absolute convergence implies regular convergence
The factors corresponding to $1 \pm a_n, |a_n| \ge 1/2$ are finitely many and non zero, so do not affect convergence or result non-zero for $Q$ versus $Q_1$