Let $\mathcal{L}_{\omega_1 \omega}$ be infinitary first-order logic (i.e. first-order logic with countable disjunctions and conjunctions), and let $\mathcal{L}_{II}^w$ be 'weak' second-order logic, i.e. second-order logic where the second-order quantifiers are interpreted as ranging over only finite subsets and relations of the domain of a structure.
I have proved that $\mathcal{L}_{\omega_1 \omega}$ is at least as expressive as $\mathcal{L}_{II}^w$ (written: $\mathcal{L}_{II}^w \leq \mathcal{L}_{\omega_1 \omega}$), in the sense that if $S$ is any symbol set and $\varphi$ is any $\mathcal{L}_{II}^w(S)$-sentence, then there is a $\mathcal{L}_{\omega_1 \omega}(S)$-sentence $\varphi'$ with the same models: i.e. if $\mathfrak{A}$ is any $S$-structure, then $\mathfrak{A} \models_w \varphi$ iff $\mathfrak{A} \models \varphi'$ (where $\models_w$ is the satisfaction relation for $\mathcal{L}_{II}^w$).
I am now wondering whether the converse is also true: i.e. are $\mathcal{L}_{\omega_1 \omega}$ and $\mathcal{L}_{II}^w$ equally expressive, or is $\mathcal{L}_{\omega_1 \omega}$ strictly more expressive than $\mathcal{L}_{II}^w$? I have a very weak intuition that $\mathcal{L}_{\omega_1 \omega}$ will be strictly more expressive, because it seems intuitively false that a countably infinite disjunction of weak second-order formulas will (in general) be equivalent to a single weak second-order formula. However, I do not yet see how to prove this (if it is even true).
One difficulty in deciding whether $\mathcal{L}_{\omega_1 \omega} \leq \mathcal{L}_{II}^w$ is that both logics satisfy (or fail to satisfy) many of the same properties: in particular, they both satisfy the downward Lowenheim-Skolem theorem (for individual sentences), and they both fail to satisfy the compactness theorem. I have yet to personally come across a significant property that distinguishes them.
If a solution to the problem of whether $\mathcal{L}_{\omega_1 \omega} \leq \mathcal{L}_{II}^w$ is known, I would greatly appreciate a hint (rather than a full solution, unless the solution is too complex or unintuitive to be easily reached by just a hint). Thanks!