1. 假设$\{r_n\}$为正项数列，且$r_n\geq3$(此条件可以用大于等于0替代)。在文章中声称，如果$\frac{r_1+r_2\cdots+r_n}{n}$是有界的，则$\limsup _ { n\to \infty } \frac { r _ { 1 } + \cdots + r _ { n } } { r _ { 1 } + \cdots + r _ { n - 1 } }=1;$
2. 类似地，如果$\frac{r_1+r_2\cdots+r_n}{n}$是无界的，则 $\limsup _ { n\to \infty } \frac { r _ { 1 } + \cdots + r _ { n }-1 } {\log_2(r_n)+n+ r _ { 1 } + \cdots + r _ { n - 1 } }=\limsup _ { n\to \infty } \frac { r _ { 1 } + \cdots + r _ { n } } { r _ { 1 } + \cdots + r _ { n - 1 } }.$

"Proof". $A_n:=\frac{r_1+\cdots+r_n}{n}$ is bounded $\iff$$r_n is also bounded. WLOG, suppose r_n\leq C. \limsup \frac{r_1+\cdots+r_n}{r_1+\cdots+r_{n-1}}=1+\limsup \frac{r_n}{r_1+\cdots+r_{n-1}}. Since \limsup r_n\geq0, \limsup \frac{1}{r_1+\cdots+r_{n-1}}\geq 0, \limsup \frac{r_n}{r_1+\cdots+r_{n-1}}\leq \limsup r_n \limsup \frac{1}{r_1+\cdots+r_{n-1}}\leq C\cdot\limsup \frac{1}{r_1+\cdots+r_{n-1}}. Recall r_n\geq 3, then \limsup \frac{1}{r_1+\cdots+r_{n-1}}\leq \limsup\frac{1}{3^{n-1}}=0, Hence, \limsup \frac{r_1+\cdots+r_n}{r_1+\cdots+r_{n-1}}=1. 上述"证明"的错误在于第一步，前n项和是有界并不能保证r_n也是有界的。和朋友LT以及SC讨论以后，发现我证明的第一行的结论存在反例。$$ r_n=\begin{cases} 3 & n\neq 2^i \\ 3+2^i & n=2^i\end{cases}, i\in \mathbb N. \qquad(1)$$显然$\frac{r_1+\cdots+r_n}{n}\leq \frac{3n+2^{i+1}-1}{n}\leq 5 (n=2^i),$但$r_n$存在一个单调无界的子列。 类似地，ZY也给出了一个似乎更加有说服力的"错误"证明（但也是在第一步出错）： It suffices to prove$C:=\limsup \frac{r_n}{r_1 +\cdots+r_{n-1}}=0.$Assume$C>0,$then$\frac{r_n}{r_1 +\cdots+r_{n-1}}>\frac{C}{2}$,$\forall n\gg 0$, i.e.$r_n\gt \frac{C}{2}(r_1 +\cdots+r_{n-1})$.$b_n:=\frac{r_1 +\cdots+r_n}{n}>\frac{r_1+\cdots+r_{n-1}+\frac{C}{2}(r_1+\cdots+r_{n-1})}{n}=(\frac{C}{2}+1)\frac{r_1 +\cdots+r_{n-1}}{n}=\frac{C+2}{2}\frac{n-1}{n} b_{n-1}, \forall n\gg 0$. Hence,$b_n> (\frac{C+2}{2})\frac{n-1}{n} b_{n-1}> (\frac{C+2}{2})^2 \frac{n-1}{n} \frac{n-2}{n-1} b_{n-2}>\cdots >(\frac{C+2}{2})^{n-N}\frac{N}{n} b_N.b_n$is not bounded. Contradiction! 由此，我不由得怀疑这些claims是错的，并试图联系原作者Uzi。Uzi恰好是我读博学校的老师。因此当天晚上，我给他写信询问情况。大约过了几个小时，Uzi给我回信， ... When writing this paper I had in mind the construction with$r_n=[\beta^n]$. Consequently, all the "calculus" lemmas were written with a hidden assumption in mind, that the sequence r_n is weakly increasing, or perhaps that (r_n) is unbounded. I failed to mention this in the paper. As stated, the claims in the paper are false. Fortunately, this does not affect the construction. ... 也就是说，他认为这些 Calculus 论断都是基于脑中$r_n$的例子给出来的，因此本身是错的。但是其构造只依赖于这一具体的$r_n$例子，是正确的。 事实上，稍作计算可以可发现，上述反例$(1)$也给出了一个论断1的反例。 Interestingly, Krause and Lenagen decided to include this example in the revised version of "Growth of Algebras and GK dimension", which came out not long after this paper. They also caught my calculus omission... we had a short exchange on this, and I send them a corrected statement and proof, which is probably in the book now. .... Uzi指出写经典 GK教材的作者 Krause 和 Lenagen在20年前就发现了这个问题，当时他做了修正，并给出了新形式的定理，收录在KL的书[2]中。Uzi在信中还和我讲了一些关于 Simple algebra 的构造。 随后加入讨论的 Beeri 给我讲解了Simple algebra的构造思路： If W is uniformly recurrent, then$A_W$is projectively simple primitive. Nekrashevych observed that in such a setting, a suitable localization of A_W (you basically invert the sum of all letters) is simple with GK-dimension given by the complexity of W. 实际上 Uzi 的构造中的$W$也是uniformaly recurrent的，这一结论Uzi没有直接指出。但是从$A_W$的prime 和 just infinite可以得到$W$就是uniformly recurrent. 因此，在某种程度上将complexity为$n^r\$的uniformly recurrent字最早就归功于 Uzi的构造。

1. U. Vishne, Primitive Algebras with Arbitrary Gelfand-Kirillov Dimension, Journal of Algebra. 211 (1999) 150–158. (pdf)
2. Krause and Lenagan, Growth of Algebras and Gelfand-Kirillov Dimension, GSM22 (2000) 162-163. (Link:P141)
3. P. Kurka, Topological and symbolic dynamics , Société Mathématique De France Paris (2003), pp.199.
4. V. Nekrashevych, Growth of étale groupoids and simple algebras , International Journal of Algebra and Computation 26 (2016), pp. 375–397.

2022-06-10