TITLE: A question about groups of intermediate growth
QUESTION [11 upvotes]: 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:

$\zeta_k(G,d_S)=1$ for all $k$ (well, $k=1$ is enough) implies that $G$ has sub-exponential growth.
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

REPLY [9 votes]: 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.