On convergence exponent

Let $0<r_1\le r_2\le\cdots$ be a real sequence satisfying $\lim\limits_{n\to \infty}r_n=+\infty$. The convergence exponent of the sequence is defined as follows. 

$$\lambda=\inf\left\{\alpha: \sum_{n=1}^\infty \frac{1}{r_n^\alpha}<\infty\right\}.$$

The convergence exponent has the following characterization:

$$\lambda=\limsup\limits_{n\to \infty}\frac{\log n}{\log r_n}.$$

Proof. Let $\beta:=\limsup\limits_{n\to \infty}\frac{\log n}{\log r_n}$ and $\alpha>\beta.$ Then there is an $\varepsilon>0$ such that $\alpha>(1+\varepsilon)\beta$. It follows from the definition of limsup there is an $N$ such that

$$\frac{\log n}{\log r_n}<\frac{\alpha}{1+\varepsilon}$$

for all $n>N$. Then , for all $n>N$, we have

$$\frac{1}{r_n^\alpha}<\frac{1}{n^{1+\varepsilon}}.$$

That is, $\sum r_n^{-\alpha}<\infty$ for all $\alpha>\beta$. Therefore, $\lambda\le \beta.$

On the other hand, fix $\alpha<\beta$, it is sufficient to show that $\sum r_n^{-\alpha}<\infty.$ By the definition of limsup, there exists an infinite sequence $\{n_k\}_{k\in N}$ such that $n_{k+1}\ge 2n_k$ and

$$\frac{\log n_k}{\log r_{n_k}}>\alpha.$$

Then we have

$$\frac{1}{r_{n_k}^\alpha}>\frac{1}{n_k},$$

and by the monotonicity assumption we have

$$\sum_{n=n_{k-1}}^{n_k}\frac{1}{r_{n}^\alpha}\ge (n_k-n_{k-1})\frac{1}{r_{n_k}^\alpha}>\frac{n_k-n_{k-1}}{n_k}\ge \frac{1}{2},$$

which shows that $\sum r_n^{-\alpha}=\infty $ for $\alpha<\beta.$ Therefore, $\lambda\ge \beta.$