Let $G$ be a finitely generated group, $S$ a fixed symmetric generating set and $B(n)$ the ball of radius $n$ about the identity with respect to the word length induced by $S$ on $G$.


Fix $k\geq1$ and denote by $\zeta\_k(G,d\_S)$ the infimum over $n\geq1$ of $\frac{|B(nk+k)|}{|B(nk)|}$.


Observe that:


1. $\zeta\_k(G,d\_S)=1$ for all $k$ (well, $k=1$ is enough) implies that $G$ has sub-exponential growth.
2. If $G$ has polynomial growth, then $\zeta\_k(G,d\_S)=1$, for all $k$ (Gromov + Pansu - by the way, is there a direct proof of this fact, without using such a big theorems?).


What happens in the middle? More formally:



> 
> **Question:** What can we say about $\zeta\_k(G,d\_S)$ when $G$ has intermediate growth? Is it always $1$? Is it always $>1$? Can be both?
> 
> 
> 


**Update:** The answer has been provided by Martin Kassabov below: the condition $\zeta\_k(G)=1$ is equivalent for $G$ to have sub-exponential growth.


Thanks in advance,


Valerio



A question about groups of intermediate growth
If $G$ has sub exponential growth one has that $\lim\_s \sqrt[s]{B(s)}=1$
if you assume that $\zeta\_k(G) = c >1$ then by an easy induction 
we have $B(nk) > K c^n$ which implies that
$$
\limsup\_s \sqrt[s]{B(s)} > \limsup\_n \sqrt[nk]{B(nk)} > \limsup\_n \sqrt[nk]{kc^n} = 
\sqrt[k]{c} > 1
$$


Therefore $\zeta\_k(G) \leq 1$, but it is clear that $\zeta\_k(G)\geq 1$, i.e
$\zeta\_k(G)=1$.


I.e. for any group of sub exponential growth $\zeta\_k(G) =1$.


The same is true if you replace the infimum in the the definition of $\zeta\_k(G)$
with limsup, but the argument is more involved and uses that sub-multiplicaticity estimates. 


