Let $G$ be a f.g. group and $d$ be a word metric w.r.t. a symmetric generating set. For $g\in G$, define $|g|:=d(g,e)$, where $e$ is the group identity. For $k\in\mathbb N$, put $$n_k:=\#\{g\in G: |g|\leq k\}\quad\text{and}\quad m_k:=\#\{g\in G: |g|=k\}$$ In which groups $\lim_k\frac{m_k}{n_k}=0$?
I don't know if it is helpful or not; all groups I work with, are torsion free.
The answer was basically given by @duh in his/her comment. The group satisfies $\frac{m_k}{n_k}\to 0$ if and only if the balls are Foelner sequences. In particular, the group has to be amenable.
Now the question becomes which amenable groups satisfy that balls are Foelner sequences ?
As suggested by @duh, virtually nilpotent groups do satisfy this condition, but you do not need to mention Foelner sequences. We know that virtually nilpotent have polynomial growth. Usually, this is stated as $a_1k^d-b_1\leq n_k\leq a_2k^d+b_2$ for some $a_1,b_1,a_2,b_2$, where $d$ is the homogeneous dimension of the group. However, Pansu improved this result, showing that $n_k\sim ck^d$ for some constant $c$. In particular, $\frac{n_k}{n_{k-1}}$ converges to 1 and since $m_k=n_k-n_{k-1}$, we indeed have that $\frac{m_k}{n_k}$ converges to 0.
