A well known convergence criterion for infinite products $\prod(1+a_n)$ is that they convergence if $\sum {a_n}$ converges, for real, not complex, $a_n\geq0$.
Proofs commonly use the following inequality: $$\prod \left( 1+a_n \right) \leq \prod \left( e^{a_n} \right) = e^{\sum a_n}$$
Question: where does the requirement for $a_n \geq 0$ come from?
The inequality used here $1+a_n \leq e^{a_n}$ is valid for all $a_n$, not just $a_n \geq 0$.
An example of where $a_n \geq0$ is stated is wikipedia's article on infinite products.
A sequence $p_1,p_2,\dots$ with an upper bound is not necessarily convergent, but if the sequence is increasing, then having an upper bound is enough to ensure it is convergent.
When all $a_i\geq0,$ then $p_n=\prod_{i=1}^n (1+a_i)$ is increasing. It is not if some of the $a_i$ are negative.
Additionally, you don’t get $$e^{\sum_{i=1}^{\infty} a_i}$$ as an upper bound for the finite products, because the finite products of $e^{a_i}$ are not increasing.
Rather, the upper bound is:
$$M=\sup_n \,e^{\sum_{k=1}^{n} a_k},$$ $M<+\infty$ because $\sum a_k$ converges, though, so we do have an upper bound.