On convergence exponent

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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 convergence exponent

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原文地址：http://www.cnblogs.com/jinjun/p/5505473.html