Are there asymptotically more nonabelian groups of order $p^k$ than there are abelian groups of order $\leq p^k$? Let $\alpha(n)$ denote the number of isomorphism classes of abelian groups of order $n$ and $\alpha^\prime(n)=\sum_{m=1}^n\alpha(m)$.  Similarly, define $f(p^k)$ to be the number of isomorphism classes of $p$-groups of order $p^k$, $\mu(p^k):=f(p^k)-\alpha(p^k)$, and $\mu^\prime(p^k)=\sum_{m=1}^k\mu(p^m)$.

Can we describe the asymptotic behavior of $$\frac{\mu(p^k)}{\alpha^\prime(p^k)}\hspace{10pt}?$$  Does it diverge? If it converges, is the limit $>1$?  If not, does it do so when we replace $\mu$ by $\mu^\prime$?

By a theorem of Higman, a lower bound for the number of $p$-groups of order $p^k$ is $$p^{\tfrac{2}{27}k^2(k-6)}\leq f(p^k),$$
which is pretty big.  Does anyone know of a good enough upper bound for $\alpha^\prime$ to be trumped by $\mu$?
 A: Let $p(t)$ denote the number of partitions of $t$. Then, for $n = \prod\limits_{i=1}^r p_i^{e_i}$, the number of isomorphism classes of abelian groups of order $n$ is
$$\alpha(n) = \prod_{i=1}^r p(e_i).$$
Now, since $r = \omega(n) \leqslant \log_2 n$ and also $e_i \leqslant \log_2 n$, a brutal estimate shows
$$\alpha(n) \leqslant p(\lceil\log_2 n\rceil)^{\log_2 n}.$$
Using the asymptotic formula for the number of partitions
$$p(t) \sim \frac{1}{4t\sqrt{3}}\exp\left(\pi \sqrt{\frac{2t}{3}}\right),$$
we can estimate
$$\alpha(n) \leqslant \exp\left(\pi \log_2 n \sqrt{\frac{2\lceil\log_2 n\rceil}{3}}\right) \leqslant n^{c\sqrt{\log n}},$$
and, since the estimate is increasing in $n$,
$$\alpha'(n) \leqslant n^{1+c\sqrt{\log n}}.$$
Inserting $n = p^k$ into that estimate yields
$$\alpha'(p^k) \leqslant p^{k(1+ c\sqrt{k\log p})} \leqslant p^{\tilde{c}k\sqrt{k}}.$$
Hence we have, for large $k$
$$\frac{\mu(p^k)}{\alpha'(p^k)} \geqslant p^{\frac{2}{27}k^2(k-6) - \tilde{c}k^{3/2}} \to \infty.$$
For large $k$, $\alpha'(p^k)$ is insignificantly small compared to $\mu(p^k)$.
