Is there a beth number larger than $\aleph_\omega$? Assuming that the GCH is false, then could there possibly exist an $n \in \mathbb N$ such that $\beth_n > \aleph_\omega$?
This would mean that the space between aleph numbers decreases more and more such that $\lim_{n \to \infty} \aleph_n = \aleph_\omega$ converges to a number on the "beth number number line" ($\aleph_\omega < \beth_\omega$).
 A: The answer is negative.
For example, if $\sf GCH$ holds below $\aleph_\omega$, with the exception that $2^{\aleph_0}=\aleph_2$, e.g. in Cohen's model of $\lnot\sf CH$, then $\beth_\omega=\aleph_\omega$, and $\beth_n=\aleph_{n+1}$ for $n>0$.
Moreover, given any function $f\colon\omega\to\omega$ which describes a consistent behavior of the continuum function below $\aleph_\omega$, it is consistent $f$ describes the continuum function below $\aleph_\omega$ and that $\beth_\omega=\aleph_\omega$.
At the same time, it is consistent that $\beth_1>\aleph_\omega$, so of course we cannot prove that $\beth_\omega=\aleph_\omega$ from $\sf ZFC$.
In terms of "space", the space between $\aleph$ numbers is considered to be constant, whereas the $\beth$ numbers can be spaced out further. The reason is that we can always increase $\beth$ numbers without collapsing cardinals. This is impossible for $\aleph$ numbers.
So to sum up, then, it is consistent that $\aleph_\omega=\beth_\omega$ along with many different behaviors of the continuum functions below $\aleph_\omega$ (and thus different $\beth$ assignments). And it is also consistent that $\beth_1>\aleph_\omega$.
Let me also remark that both $\aleph$ and $\beth$ numbers are indexed in terms of ordinals, and they certainly continue well after $\omega$. There is a proper class of $\aleph$ and $\beth$ numbers.
