Let $\Gamma$ be a finitely generated group of exponential growth and $gr(S)=\lim\_{k\rightarrow \infty} \sqrt[k]{|B\_k(S)|}$ be the growth rate of $\Gamma$ with respect to the generating set $S$. I am confused with the following question: Does there always exist a generating set $S'$ such that
$$\frac{|B\_k(S')|}{gr(S')^k}\rightarrow 1, \text{ when }k\rightarrow \infty$$



Asymptotics of the growth rate of a group
There is never a (finite) generating set with that property.


Consider a generating set $S=\{x\_1,\ldots,x\_{\ell}\}$ of cardinality $\ell$. Let $B\_k := B\_k(S)$, $S\_k := B\_k \setminus B\_{k-1}$, and $g := gr(S)$. Let $b\_k := |B\_k|$ and $s\_k: = |S\_k|$. Assume for simplicity that $L := \lim\_{k \to \infty} \frac{b\_k}{g^k}$ exists (although it shouldn't be difficult to get general liminf bounds). Also set
$$ L' := \lim\_{k \to \infty} \frac{s\_k}{g^k} = \lim\_{k \to \infty} \frac{b\_k-b\_{k-1}}{g^k} = \left(1-\frac{1}{g}\right)L. $$


For the spheres, there's a trivial inequality $s\_{m+n} \leq s\_m s\_n$. (Any word of minimal length $m+n$ in the generators may be written at least one way as a product of two words of minimal length $m$ and $n$.) This is already enough to give $L' \geq 1$, and thus
$$ L \geq \frac{1}{1-\frac{1}{g}} > 1. $$


This is unsatisfying and still leaves the possibility that $L'=1$. Thus, I eliminate that possibility as well, by improving the trivial bound $s\_{2k} \leq s\_k^2$ to $s\_{2k} \leq \left(1-\frac{1}{2\ell}\right)s\_k^2$. This improvement (unlike the trivial improvement above) does not hold for monoids, so the extra cancellation comes (unsurprisingly) from the existance of inverses.


Specifically, fix $k$, let $E\_i$ be the set of all elements of $S\_k$ which may be written as a word of length $k$ ending in $x\_i$, and let $E\_{\ell+i}$ be the set of all elements of $S\_k$ which may be written as a word of length $k$ ending in $x\_i^{-1}$. Set
$$ F\_i = E\_i \setminus \bigcup\_{j < i} E\_j, $$
so the $F\_i$ are disjoint.
Let $F\_i'$ be the image of $F\_i$ under the inversion map, and let $n\_i = |F\_i|$. The product of an element of $F\_i$ and an element of $F\_i'$ may be written with $2k-2$ letters (because the $x\_i$ and $x\_i^{-1}$ cancel), so those $n\_i^2$ products are not in $S\_{2k}$. Thus, we have
$$ s\_{2k} \leq s\_k^2 - \sum\_{i=1}^{2\ell} n\_i^2. $$
By the Hölder inequality, we have
$$ s\_k = \sum\_{i=1}^{2\ell} n\_i \leq \sqrt{\sum\_{i=1}^{2\ell} 1^2} \sqrt{\sum\_{i=1}^{2\ell} n\_i^2} $$
$$ \frac{s\_k^2}{2\ell} \leq \sum\_{i=1}^{2\ell} n\_i^2. $$
Combining the inequalities gives
$$ s\_{2k} \leq \left(1-\frac{1}{2\ell}\right)s\_k^2. $$
Dividing by $g^{2k}$ and sending $k \to \infty$ gives
$$ L' \geq \frac{1}{1-\frac{1}{2\ell}} > 1. $$
For balls, we obtain the bound
$$ L \geq \frac{1}{\left(1-\frac{1}{g}\right)\left(1-\frac{1}{2\ell}\right)}. $$
This bound is optimal for free generating sets of free groups.


I'd be interested in seeing what Kate's limit says (or don't say) about the underlying group. Can the lower bound I just gave be achieved for non-free groups? Can the limit generally be made arbitrarily close to $1$ by increasing the number of generators appropriately (as Misha speculated in the comments)? I'm far from an expert in combinatorial/asymptotic group theory, so I don't have good intuition for what intrinsic information this value holds.


