Generating series - Finite groups of order $n$ I am wondering if something of interest can be said about one of the two series
$$G_1(x)=\sum_{n=1}^{+\infty}{\mathcal{G}(n)z^n}$$
$$G_2(s)=\sum_{n=1}^{+\infty}{\frac{\mathcal{G}(n)}{n^s}}$$
where $\mathcal{G}(n)$ is the number of finite groups of order $n$. (I don't even know if those series are converging for any $z$ or $s$...)
Thanks a lot !
 A: Some things can be said about this, you should take a look at the book "$\underline{\text{Enumeration of finite groups}}$" by Blackburn, Neumann and Venkataraman. In this book, if for any $n$, $\mu(n)$ denotes $max(\alpha_i)$ where :
$$n=\prod_{i=1}^rp_i^{\alpha_i} $$
Then you have the following upper bound proven by Laszlo Piber proven in 1991 :
$$\mathcal{G}(n)\leq n^{\frac{2}{27}\mu(n)^2+O(\mu(n)^{\frac{5}{3}})} $$
Now, if you want to talk convergence of $G_1$ I think it is possible, take $n$ and $k$ such that :
$$2^k\leq n<2^{k+1} $$
It is somehow clear that $\mu(n)<k+1$ (otherwise $n$ would be divisible by $p^{k+1}\geq 2^{k+1}$), on the other hand :
$$k\leq log_2(n)<k+1$$
Hence we have that :
$$\mu(n)<log_2(n)+1 $$
Now I would like to find a good majoration for the upper bound. You have that :
$$(n^{\frac{2}{27}\mu(n)^2+O(\mu(n)^{\frac{5}{3}})})^{\frac{1}{(log_2(n)+1)^2}}\leq n^K$$
For some positive constant $K$. Hence :
$$\mathcal{G}(n)\leq n^{K(log_2(n)+1)^2}$$
Now you have a criterion for convergence of such series (I don't remember the name) if you positive terms $a_n$ for a serie with $R$ its radius of convergence then if :
$$\sqrt[n]{a_n} \text{ has a limit }L$$
Then $L=\frac{1}{R}$ but if $a_n= n^{K(log_2(n)+1)^2}$ then :
$$\sqrt[n]{a_n}=exp(K(log_2(n)+1)^2log(n)\frac{1}{n})$$
This is easily seen to be convergent to $1$ hence the radius for the serie associated to $(n^{K(log_2(n)+1)^2})$ is $1$ and for the serie  $G_1$ it is greater than $1$ (by comparison theorem). On the other hand it is more than clear that $\mathcal{G}(n)$ being $\geq 1$ and hence that the radius of convergence for $G_1$ will be $\leq 1$. 
Hence the radius of convergence of $G_1$ is $1$. 
For $G_2$ I think it is nowhere convergent. I will actually show that there are no real numbers where it is convergent. First, because $\mathcal{G}(n)\geq 1$ you have that if $G_2(s)$ converge then $s>1$. Now take $s>0$, one thing you know about lower bound of $\mathcal{G}(n)$ is the following :
$$\mathcal{G}(p^k)\geq p^{\frac{2}{27}k^3-O(k^2)} $$
Now if $N>p^m$ then :
$$\sum_{n=1}^N\frac{\mathcal{G}(n)}{n^s}\geq \sum_{k=1}^m\frac{\mathcal{G}(p^k)}{p^{ks}} \geq \sum_{k=1}^m p^{\frac{2}{27}k^3-O(k^2)-ks}$$
Now $O(k^2)-ks=O(k^2)$ hence :
$$\sum_{n=1}^N\frac{\mathcal{G}(n)}{n^s}\geq\sum_{k=1}^m p^{\frac{2}{27}k^3-O(k^2)}\geq \sum_{k=1}^m p^{\frac{2}{27}k^3-C_sk^2}$$
Which is easily seen to be converging (when $m$ goes to infinity) to infinity. Hence the sum cannot converge when $N$ goes to infinity hence there cannot be any convergence... 
