# balls as Foelner sets

This is essentially equivalent to this question by Simon Thomas. Let $G=\langle X\rangle$ be a finitely generated group, $b_n$ be the number of elements in the ball of radius $n$ in the Cayley graph.

1) Is it possible that the limit $\lim \frac{b_{n+1}}{b_n}$ does not exist?

2) Suppose that for every constant $\epsilon>0$ there exists an $n$ such that $\frac{b_{n+1}}{b_n}\le 1+\epsilon$. Does it imply that $\lim \frac{b_{n+1}}{b_n}=1$?

Note that the condition of 2) implies that the group is amenable and one can take balls as Foelner sets (which would contradict a statement in de la Harpe's book).

Update: The first part has been asked and answered already before (see Andreas' answer below). About 2): here is a stronger question. Suppose that an amenable group $G$ is finitely presented. Are there constants $\epsilon>0, N$ depending only on the lengths of the defining relations so that if $\frac{b_{n+1}}{b_n}\le 1+\epsilon$ for some $n>N$, then the limit above exists and is equal to 1? This is similar to a statement proved by Shalom and Tao about polynomial growth, but for groups of subexponential growth.

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I'm voting to close this as a duplicate. –  HJRW Oct 28 '10 at 16:11
A post on meta.MO would probably drum up the necessary votes. –  HJRW Oct 28 '10 at 17:20
Ah, yes, your new question is indeed presumably not a duplicate! I don't know if I can undo my vote, but I'll try... –  HJRW Oct 28 '10 at 23:38

@Andreas: Question 2 is not exactly the same as Simon's. It might be possible that if for some $n$ $b_{n+1}/b_n<1+\epsilon$ for some very small (universal constant, depending on the presentation) $\epsilon$, then the limit exists and is equal to 1. That is why I formulated it this way. The answer could be different for finitely presented and infinitely presented groups. –  Mark Sapir Oct 28 '10 at 16:41