Let $\{q_n\}$ be a sequence of positive integers satisfying the condition that $q_n\geq 2$ ($n=1, 2, \cdots$), then every real number $x\in[0, 1]$ can be represented as a Cantor series

$ x=\sum\limits_{n=1}^{\infty}\frac{\varepsilon_n(x)}{q_1q_2\cdots q_n}, $

where $\varepsilon_n(x)\in\{0, 1, \cdots, q_n-1\}$ is called the $n$-th digit in the Cantor expansion of $x$.

Note that in the special case of $q_n=q$ for a fixed integer $q\geq 2$, the Cantor series corresponds to the $q$-ary expansion of $x$.

In Galambos' book [2], there is a good account of the Cantor expansion and many other representations of real numbers by infinite series, from different viewpoints such as probability theory as well as metric number theory.

In [1], several interesting statistical properties of the digit sequence $\{\varepsilon_n(x)\}$ were investigated by Erdös and Renyi. For instance, let $d_n(x)$ denote the number of distinct numbers in the first $n$ digits $\{\varepsilon_1(x), \varepsilon_2(x), \cdots, \varepsilon_n(x)\}$ of the Cantor series, it was proved that

$ \lim\limits_{n\to\infty}\frac{d_n(x)}{n}=1 $

for almost all $x$ (with respect to the Lebesgue measure), under the condition that

$ \sum\limits_{n=1}^{\infty}\frac{1}{q_n}<\infty. $

In this note, we are concerned with the following set

$ E_{\delta}=\{x\in[0, 1]:\lim\limits_{n\to\infty}\frac{d_n(x)}{n^{\delta}}=c\}, $

where $0<\delta<1$ and $c>0$. By the theorem of Erdös and Renyi, the set $E_{\delta}$ has Lebesgue measure zero. We shall prove the following

Theorem 1   The set $E_{\delta}$ has Hausdorff dimension $\delta$ provided that

$ \lim\limits_{n\to\infty}q_n=\infty\textrm{ and }\lim\limits_{n\to\infty}\frac{\log q_{n+1}}{\log q_n}=1. $

Let us remark that if $\{q_n\}$ increases rapidly, then the set $E_{\delta}$ always has Hausdorff dimension $1$.

In order to avoid complicated calculation, we only present a proof for the special case that $q_n=n+1$ and $c=1$.

Let us begin with some notations. For each $n\geq 1$, define

$ I(a_1, a_2, \cdots, a_n)=\{x\in[0, 1]:\varepsilon_1(x)=a_1, \varepsilon_2(x)=a_2, \cdots, \varepsilon_n(x)=a_n\}, $

which is an interval of length $|I(a_1, a_2, \cdots, a_n)|=\frac{1}{(n+1)!}$. We call it a a rank-$n$ basic interval.

First we bound the Hausdorff dimension $\dim_H(E_{\delta})$ from above. For each $n\geq 2$, the set $E_{\delta}$ can be covered by nearly

$ N(E_{\delta}, n):={n\choose[n^{\delta}]}[n^{\delta}]![n^{\delta}]^{n-[n^{\delta}]} $

rank-$n$ basic intervals. It follows from Stirling's formula that

$ \dim_H(E_{\delta})\leq \lim\limits_{n\to\infty}\frac{\log N(E_{\delta}, n)}{\log(n+1)!}=\delta. $

Actually, the right hand side is an upper bound for the box-counting dimension of the set $\dim_H(E_{\delta})$. For more details about the Hausdorf and box-counting dimension, we refer to Falconer's book [3].

To bound the Hausdorff dimension $\dim_H(E_{\delta})$ from below, we construct a subset of $E_{\delta}$ as follows. Let $F_{\delta}$ be the set of $x\in[0, 1]$ subject to the restriction that

$ \varepsilon_n(x)\in\{1, 2, \cdots, [n^{\delta}]\}, \quad n=1, 2, \cdots. $

The set $F_{\delta}$ is a homogeneous Moran set (see [4]) which has Hausdorff dimension

$ \dim_H(F_{\delta})=\lim\limits_{n\to\infty}\frac{\log \big(\prod\limits_{k=1}^{n}[k^{\delta}]\big)}{\log(n+1)!}=\delta. $

It is easy to see that $F_{\delta}\subset E_{\delta} $. The proof is finished now.

Acknowledgement  The author is grateful to Professor Jihua Ma for suggesting this problem.