Theorem A. Suppose $a_n > 0$. Then $\sum a_n < \infty$ iff $\prod (1 + a_n) < \infty$.
Proof. The partial products are clearly increasing. Also note via expansion that
$$
(1 + a_1)(1 + a_2)\cdots(1 + a_n) \leq e^{a_1 + a_2 + \ldots + a_n}
$$
since $x + 1 \leq e^x$ for all real $x$. By assumption, the right-hand side converges to $e^{\sum a_n}$, so the partial products are also bounded. Hence they converge.
Theorem B. Let $a_n \in \Bbb C$ be such that $\sum |a_n| < \infty$. Then the product $\prod (1 + a_n) < \infty$, and equals $0$ iff one of the terms $(1 + a_n)$ is zero.
Proof. If one of the terms is zero, then the product obviously converges to zero. So let’s assume the other direction.
For all sufficiently large $n$, we have $|a_n| < 1/2$. We can disregard any prior terms, since they don’t affect the convergence of the product. Let’s write
$$
\begin{eqnarray}
\prod_{k=1}^n (1 + a_k) &=& \prod_{k=1}^n e^{\ln(1 + a_k)} \\
&=& e^{\sum_{k=1}^n \ln(1 + a_k)} \\
&=& e^{\sum_{k=1}^n b_k}
\end{eqnarray}
$$
where $b_k = \ln(1 + a_k)$. Next, note that $\ln(z)$ is holomorphic on the disk centered at $1$ with radius $1/2$, and so can be expanded there to the infinite series
$$
\ln(1 + z) = z – z^2/2 + z^3/3 – \ldots
$$
which implies $|\ln(1 + z)| \leq 2|z|$ when |z| < $1/2$: for we have
$$
\begin{eqnarray}
|z^2/2 – z^3/3 + z^4/4 – \ldots| &=& |z| |z/2 – z^2/3 + z^3/4 – \ldots| \\
&\leq& |z| \sum_{k=1}^{\infty} |z^k| \\
&\leq& |z| \sum_{k=1}^{\infty} \frac{1}{2^k} \\
&\leq& |z|.
\end{eqnarray}
$$
So $\sum \ln(1 + a_n)$ must converge absolutely, by comparison to $\sum 2a_n < \infty$. Therefore the partial products above converge to $e^{\sum b_k}$, which is non-zero since $\sum b_k$ is finite.
Corollary. If $\sum |1 – a_n|$ converges, so does $\prod a_n$; and the latter converges to a non-zero number iff all the terms are non-zero.
Weierstrass Factorization
Given a countable sequence $A = \{a_i\}$, we want to find an entire function $f$ which vanishes on the $a_i$’s and nowhere else. One way to do this is to take $f = \prod_n f_n$, where $f_n(a_n) = 0$. But how can we also make it so that this product doesn’t vanish anywhere else?
The above Corollary above gives us a hint: if we can force $\sum |1 – f_n(z)|$ to converge, then so will $\prod f_n(z)$ to some non-zero number.
Note that via the expansion $\ln(1 – z) = -z + -z^2/2 – \ldots$, we have for $|z| < 1/2$
$$
\begin{eqnarray}
|1 – E_k(z)| &=& \left| 1 – \exp \left( \frac{z^{k+1}}{k+1} + \frac{z^{k+2}}{k+2} + \ldots \right) \right| \\
&=& \left| 1 – \exp (w) \right|.
\end{eqnarray}
$$
But $|z| < 1/2$ implies $|w| \leq 2|z|^{k+1} \leq 1$. Thus $|1 – e^w| \leq c|w|$ for some constant $c$, which in turn implies $|1 – E_k(z)| \leq 2c|z|^{k+1}$. Then $\sum_k |1 – E_k(z)| < \infty$ for all $|z| < 1/2$ so $\prod_k E_k(z)$ converges.
Define $F_n = E_k(z/a_n)$. Since $|a_n| \to \infty$, for any $z$, all but finitely many of the terms $z/a_n$ have magnitude less than $1/2$; if $z \notin A$, this means they are all nonzero. Hence the zeroes are precisely the $a_i$’s.