On convergence exponent

Let $0<r_1\le r_2\le\cdots$ is 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 was an $\varepsilon> 0$ such that $\alpha> (1+\varepsilon) \beta$. It follows from the definition of limsup there are 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's, $\sum r_n^{-\alpha}<\infty$ for all $\alpha>\beta$. Therefore, $\lambda\le \beta.$

On convergence exponent