Let $G$ be a reduced abelian $p$-group. We set $G_0=G$. Let $\alpha$ be an ordinal. Inductively, if $\alpha=\beta+1$ is a successor ordinal, we define $$G_{\alpha}=pG_{\beta}.$$ If $\alpha$ is a limit ordinal, we define $$G_{\alpha}=\bigcap_{\beta<\alpha}G_{\beta}.$$ Let $P$ be the set of all elements $x\in G$ satisfying $px=0$. Then the $\alpha$-th Ulm invariant is defined as $$f(\alpha,G)=\dim(P\cap G_{\alpha})/(P\cap G_{\alpha+1}).$$ I am trying to solve Exercise 44 on Kaplansky's book ''Infinite Abelian Group'':

If for every finite $n$, its Ulm invariant satisfies $f(n)\leqslant\aleph_0$, then $f(\omega)\leqslant\aleph_0$.

I go through Fuchs' book ''infinite abelian group''. A function $g$ from the ordinals $<\tau$ to the cardinals is said to be $\tau$-admissible if $$(1)\ \tau=\sup\{\sigma+1\mid g(\sigma)\ne 0\},$$ $$(2) \sum_{\rho\geqslant\sigma+\omega}g(\rho)\leqslant\sum_{n<\omega}g(\sigma+n), \text{ for all } \sigma \text{ with } \sigma+\omega<\tau.$$ If $\tau>\omega$, and we take $\sigma=0$, then (2) becomes $$\sum_{\rho\geqslant\omega}g(\rho)\leqslant\sum_{n<\omega}g(n).$$ As a corollary, Exercise 44 holds if $f$ is admissible. Theorem 78.6 on Fuchs' book states that:

If $G$ is a direct sum of countable groups, then $f$ is admissible.

Thus, Exercise 44 holds when $G$ is a direct sum of countable groups.

I don't know how to solve Exercise 44 for arbitrary abelian $p$-groups. Any comments and suggestions are welcome. Thank you very much for your kind help.