An infinite product converges iff its series of logarithms converges

Remarks

The partial products and the partial sums are linked by $\prod_{n \le N} a_n = \exp\left(\sum_{n \le N} \ln a_n\right)$. The map $\exp$ is a continuous bijection from $\R$ onto $(0, \infty)$, with continuous inverse $\ln$. So the partial products tend to a positive limit exactly when the partial sums tend to a real limit. The nonzero condition in the definition of convergence is what makes this work: partial sums tending to $-\infty$ correspond to products diverging to $0$. A consequence: if $u_n \ge 0$, then $\prod (1 + u_n)$ converges if and only if $\sum u_n$ converges. Indeed, if $u_n \to 0$ then $\ln(1 + u_n) / u_n \to 1$, and the limit comparison test applies; if $u_n \not\to 0$, both diverge. Example: $\prod (1 + 1/n^2)$ converges and $\prod (1 + 1/n)$ diverges.

Going deeper

For a student

The statement reduces the theory of products of positive numbers to the theory of series. Its most used form is the criterion for $\prod (1 + u_n)$ with $u_n \ge 0$, which compares the product with $\sum u_n$. For complex factors one uses a branch of the logarithm near $1$, and absolute convergence of $\sum (a_n - 1)$ gives convergence of the product.