[裴礼文数学分析中的典型问题与方法习题参考解答]5.1.11

证明: 若 $a_n>0$, $a_n\searrow 0$, 则 $\dps{\vsm{n}a_n}$ 与 $\dps{\vsm{m}p_m2^{-m}}$ ($p_m=\max\sed{n;a_n\geq 2^{-m}}$) 同时敛散. (Lobachevsky 判别法)

证明: 由 $$\beex \bea \sum_{k=1}^m p_k2^{-k} &=\sum_{k=1}^m p_k\sex{\frac{1}{2^{k-1}}-\frac{1}{2^k}}\\ &=\sum_{k=0}^{m-1}\frac{p_{k+1}}{2^k} -\sum_{k=1}^m \frac{p_k}{2^k}\\ &=p_0-\frac{p_m}{2^m} +\sum_{k=1}^{m-1}\frac{p_{k+1}-p_k}{2^k}\\ &=p_0-\frac{p_m}{2^m} +2\sum_{k=1}^{m-1}(p_{k+1}-p_k)2^{-(k+1)}\\ &\leq p_0+2\sum_{k=1}^{m-1} (p_{k+1}-p_k)a_{p_{k+1}}\\ &\leq p_0+2 \sum_{k=1}^{m-1}(a_{p_k+1}+\cdots+a_{p_{k+1}})\\ &=p_0+2\sex{a_{p_1+1}+a_{p_1+2}+\cdots+a_{p_m}} \eea \eeex$$ 知若 $\dps{\vsm{n}a_n}$ 收敛, 则 $\dps{\vsm{m}p_m2^{-m}}$ 也收敛; 若 $\dps{\vsm{m}p_m2^{-m}}$ 发散, 则 $\dps{\vsm{n}a_n}$ 也发散. 又由 $$\beex \bea \sum_{k=1}^m p_k2^{-k} &=\sum_{k=1}^m p_k\sex{\frac{1}{2^{k-1}}-\frac{1}{2^k}}\\ &=\sum_{k=0}^{m-1}\frac{p_{k+1}}{2^k} -\sum_{k=1}^m \frac{p_k}{2^k}\\ &=p_0-\frac{p_m}{2^m} +\sum_{k=1}^{m-1}\frac{p_{k+1}-p_k}{2^k}\\ &=p_0-\frac{p_m}{2^m} +2\sum_{k=1}^{m-1}(p_{k+1}-p_k)2^{-(k+1)}\\ &>p_0-\frac{p_m}{2^m} +2\sum_{k=1}^{m-1} (p_{k+1}-p_k)a_{p_{k+1}+1}\\ &\geq p_0-\frac{p_m}{2^m} +2\sum_{k=1}^{m-1}\sex{a_{p_k+2}+\cdots+a_{p_{k+1}+1}}\\ &=p_0-\frac{p_m}{2^m} +2\sex{a_{p_1+2}+a_{p_1+3}+\cdots+a_{p_m+1}} \eea \eeex$$ 知若 $\dps{\vsm{m}p_m2^{-m}}$ 收敛, 则 $\dps{\vsm{n}a_n}$ 也收敛; 若 $\dps{\vsm{n}a_n}$ 发散, 则$\dps{\vsm{m}p_m2^{-m}}$ 也发散.