设 $a_n>0$, $\dps{\vsm{n}a_n}$ 收敛, $na_n$ 单调, 证明: $$\bex \vlm{n}na_n\ln n=0. \eex$$
证明: 又题意, $na_n\searrow 0$. 又由 Cauchy 收敛原理, $\forall\ \ve>0,\ \exists\ N,\st n\geq N\ra$ $$\beex \bea \frac{\ve}{2}&>\sev{\sum_{k=[\sqrt{n}]}^{n-1}a_k} =\sev{\sum_{k=[\sqrt{n}]}^{n-1} \frac{1}{k}\cdot ka_k}\\ &\geq na_n \sum_{k=[\sqrt{n}]}^{n-1}\frac{1}{k} \geq na_n\sum_{k=[\sqrt{n}]}^{n-1} \int_k^{k+1}\frac{1}{x}\rd x\\ &=na_n\int_{[\sqrt{n}]}^n\frac{1}{x}\rd x\geq na_n\int_{\sqrt{n}}^n\frac{1}{x}\rd x=\frac{1}{2} na_n\ln n. \eea \eeex$$
