设 $a_n>0$ ($n=1,2,\cdots$) 且 $\dps{\vsm{n}a_n}$ 收敛, $\dps{r_n=\sum_{k=n}^\infty a_k}$. 试证:

(1). $\dps{\vsm{n}\frac{a_n}{r_n}}$ 发散.

(2). $\dps{\vsm{n}\frac{a_n}{\sqrt{r_n}}}$ 收敛.

证明:

(1). 由 $$\bex \sum_{k=n+1}^{n+p}\frac{a_k}{r_k} \geq \frac{1}{r_{n+1}}\sum_{k=n+1}^{n+p}a_k\to 1\quad\sex{p\to\infty} \eex$$ 知 $$\bex \exists\ p,\st \sum_{k=n+1}^{n+p}\frac{a_k}{r_k}\geq \frac{1}{2}. \eex$$

(2). $$\beex \bea \vsm{n}\frac{a_n}{\sqrt{r_n}}&=\vsm{n}\frac{r_n-r_{n+1}}{\sqrt{r_n}} =\vsm{n}\frac{(\sqrt{r_n})^2-(\sqrt{r_{n+1}})^2}{\sqrt{r_n}}\\ &=\vsm{n}\frac{2\sqrt{\xi_n}(\sqrt{r_n}-\sqrt{r_{n+1}})}{\sqrt{r_n}}\quad\sex{r_{n+1}<\xi_n<r_n}\\ &\leq 2 \vsm{n} \sex{\sqrt{r_n}-\sqrt{r_{n+1}}}\\ &=2\sqrt{r_1}. \eea \eeex$$