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This is essentially equivalent to this 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.

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.

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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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. IsAre there a 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.

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. Is there a 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.

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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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. Is there a 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.

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).

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. Is there a 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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user6976
user6976
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