Equivalence between beth number, $H$ and $R$ function - singular case

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I want to prove the following

For $\kappa$ infinite cardinal, $H_\kappa=R_\kappa$ if and only if $\kappa=\beth_\kappa$

With $R_\kappa$ the von Neumann universes, $H_\kappa$ be the set of all sets whose transitive closure has size less than $\kappa$. $$ H_\kappa=\{ x,\ \vert \text{tr cl}(x)\vert<\kappa$$

My attemp : I know that for an infinite cardinal $\vert R_\kappa\vert=\beth_\kappa$, and that $H_\kappa\subseteq V_\kappa$.

I think I can prove by induction that for $\kappa$ regular, $H_\kappa=R_\kappa$ if and only if $\kappa$ is either $\omega$ or strongly inaccessible. Therefore for $\kappa$ regular, $H_\kappa=R_\kappa$ if and only if $\vert R_\kappa\vert=\kappa$. proving my claim in the regular case.

But How can I conclude for the singular case?

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The proof actually isn't too far off for the singular case either.

Since $|R_\kappa|=\kappa$, every $x\in R_\kappa$ is in fact in some $R_\alpha$ for $\alpha<\kappa$, so it has a transitive closure of size at most $\beth_\alpha<\kappa$. Therefore $R_\kappa\subseteq H_\kappa$.

In the other direction you can still prove this by induction on $\alpha=\operatorname{rank}(x)$ for $x\in H_\kappa$, that $\alpha<\kappa$. So we get $H_\kappa\subseteq R_\kappa$.


In other words, regularity is a red herring.