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 supremum, but the argument is more involved and uses that sub-multiplicaticity estimates.

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.