The relationship between convergence exponent and upper exponential density

Posted 2020-07-28

Let $A=\{a_1<a_2<\cdots<a_n<\cdots\}\subset N$ be a integer sequence. The upper exponential density of it is defined by 

$$\overline{\varepsilon}(A)=\limsup\limits_{n\to \infty}\frac{\log|A\cap\{1,2,\cdots,n\}|}{\log n}.$$

We can similarly define the lower exponential density.

We have the following result claimed in Gaps and the exponent of convergence of an integer sequence .

Main result: $$\overline{\varepsilon}(A)=\tau(A)=\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}.$$

Proof. Let $\overline{\varepsilon}(A)<\beta.$ For any $\varepsilon>0,$ by the definition of limsup there exists an integer $N$ such that $|A\cap\{1,2,\cdots,n\}|<n^{\beta+\varepsilon}$ for any $n>N.$ It follows that $a_{[n^{\beta+\varepsilon}]}\le n$. Therefore,

$$\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}\ge \limsup\limits_{n\to \infty}\frac{\log [n^{\beta+\varepsilon}]}{\log a_{[n^{\beta+\varepsilon}]}} \ge \limsup\limits_{n\to \infty}\frac{(\beta+\varepsilon)\log n}{\log n}= \beta+\varepsilon,$$

which implies that 

$$\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}\ge\limsup\limits_{n\to \infty}\frac{\log |A\cap\{1,2,\cdots,n\}|}{\log n}.\quad (*)$$

 

On the other hand, let $\alpha<\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}$. For any $\varepsilon>0,$ there exists a sequence $\{n_k\}$ and $K$ such that $\frac{\log n_k}{\log a_k}>\alpha-\varepsilon$ for any $k>K.$ That is, $a_{n_k}<n_k^{\frac{1}{\alpha-\varepsilon}}$ (It means that $\{a_1,\cdots, a_{n_k}\}\subset \{1,2,\cdots, n_k^{\frac{1}{\alpha-\varepsilon}}\}$). Therefore, $|A\cap\{1,2,\cdots,[n_k^{\frac{1}{\alpha-\varepsilon}}]\}|\ge n_k$. So, 

$$\limsup\limits_{n\to \infty}\frac{\log |A\cap\{1,2,\cdots,n\}|}{\log n}\ge \limsup\limits_{n\to \infty}\frac{\log |A\cap\{1,2,\cdots,[n_k^{\frac{1}{\alpha-\varepsilon}}]\}|}{\log n_k^{\frac{1}{\alpha-\varepsilon}}}\ge  \limsup\limits_{n\to \infty}\frac{n_k}{\log n_k^{\frac{1}{\alpha-\varepsilon}}}=\alpha-\varepsilon, $$

which implies that

$$\limsup\limits_{n\to \infty}\frac{\log |A\cap\{1,2,\cdots,n\}|}{\log n}\ge \limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}. \quad (**)$$

Finally, the conclusion follws from the inequalities $(*)$ and $(**)$.