设数 $a>0$, $\sed{p_n}$ 是一个数列, 并且 $p_n>0$, $p_{n+1}\geq p_n$. 证明: 级数 $$\bex \vsm{n}\frac{p_n-p_{n-1}}{p_np_{n-1}^a} \eex$$ 收敛. (国外赛题)
证明: 由 $p_{n+1}\geq p_n$ 知
(1). $\dps{\vlm{n}p_n=p>0}$. 此时, $$\bex \vsm{n}\frac{p_n-p_{n-1}}{p_np_{n-1}^a} \leq \frac{1}{p_1p_0^a}\vsm{n}(p_n-p_{n-1}) =\frac{p-p_0}{p_1p_0^a}. \eex$$
(2). 或 $\dps{\vlm{n}p_n=+\infty}$. 此时, 若 $a\geq 1$, 则 $\exists\ N,\st n\geq N\ra$ $$\bex p_n\geq 1\ra \frac{p_n-p_{n-1}}{p_np_{n-1}^a} \leq \frac{p_n-p_{n-1}}{p_np_{n-1}}=\frac{1}{p_{n-1}}-\frac{1}{p_n}, \eex$$ 而 $$\bex \sum_{n=N}^\infty \frac{p_n-p_{n-1}}{p_np_{n-1}}\leq \frac{1}{p_{N-1}}. \eex$$ 若 $0<a<1$, 则 $$\beex \bea \frac{1}{p_{n-1}^a}-\frac{1}{p_n^a} &=\frac{p_n^a-p_{n-1}^a}{p_{n-1}^ap_n^a} =\frac{a\xi^{a-1} (p_n-p_{n-1})}{p_{n-1}^ap_n^a}\quad\sex{p_{n-1}<\xi<p_n}\\ &\geq \frac{ap_{n-1}^{a-1}(p_n-p_{n-1})}{p_{n-1}^ap_n^a} =a \frac{p_n-p_{n-1}}{p_np_{n-1}^a}, \eea \eeex$$ 而 $$\bex \vsm{n}\frac{p_n-p_{n-1}}{p_np_{n-1}^a}\leq \frac{1}{a}\vsm{n}\sex{\frac{1}{p_{n-1}^a}-\frac{1}{p_n^a}} =\frac{1}{ap_0^a}. \eex$$
