设 $\dps{\vsm{n}a_n}$ 收敛, $0<p_n\nearrow+\infty$, 试证: $$\bex \vlm{n} \frac{p_1a_1+p_2a_2+\cdots+p_na_n}{p_n}=0. \eex$$
证明: 设 $\dps{S_n=\sum_{k=1}^na_k}$, 则 $\dps{\vlm{n}S_n=S}$, 且 $$\beex \bea \frac{1}{p_n}\sum_{k=1}^n p_ka_k &=\frac{1}{p_n}\sum_{k=1}^n p_k(S_k-S_{k-1})\quad\sex{S_0=0}\\ &=\frac{1}{p_n}\sex{\sum_{k=1}^n p_kS_k-\sum_{k=0}^{n-1}p_{k+1}S_k}\\ &=\frac{1}{p_n}\sex{p_nS_n+\sum_{k=1}^{n-1} (p_k-p_{k+1})S_k}\\ &=S_n-\frac{1}{p_n}\sum_{k=1}^{n-1} (p_{k+1}-p_k)S_k,\\ \vlm{n}\frac{1}{p_n}\sum_{k=1}^n p_ka_k &=S-\vlm{n}\frac{1}{p_n}\sum_{k=1}^{n-1} (p_{k+1}-p_k)S_k\\ &=S-\vlm{n}\frac{(p_{n+1}-p_n)S_n}{p_{n+1}-p_n}\quad\sex{Stolz\mbox{ 公式}}\\ &=S-S=0. \eea \eeex$$
