The relationship between convergence exponent and upper exponential density

Let $A =\{a_1<a_2<\cdots<a_n<\cdots\}\subset n$ is 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 for an integer sequence.

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

Proof. Let $\overline{\varepsilon} (A) <\beta.$-$\varepsilon>0,$ 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 $\frac{\log N_k}{\log A_k}>\alpha-\vare psilon$ for any $k >k.$ that's, $a _{n_k}<n_k^{\frac{1}{\alpha-\varepsilon}}$ (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 $ (* *) $.

The relationship between convergence exponent and upper exponential density