来着幸子网友问的一个级数题

陈洪葛 posted @ Jun 09, 2014 11:12:14 AM in 数学分析 , 679 阅读

正项级数$\displaystyle\sum_{n=1}^{\infty}a_{n}$,若存在正数$\alpha$和$\beta$使得$a_{n}-a_{n+1}\geq \beta a_{n}^{2-\alpha}$,那么$\displaystyle\sum_{n=1}^{\infty}a_{n}<+\infty$,且$\displaystyle\sum_{n=N}^{\infty}a_{n}=O(a_{N}^{\alpha}),N\to\infty$.

证明:首先由
\[ a_{n}-a_{n+1}\geq \beta a_{n}^{2-\alpha}\geq 0\]
知道$a_{n}$单调递减,有下界$0$,则有
\[ \lim_{n\to\infty}a_{n}=a\geq 0 \]
现在讨论$\alpha$,若$\alpha\geq 2$,则$a_{n}^{2-\alpha}\geq a_{1}^{2-\alpha}$,得到
\[ a_{1}\geq n\beta a_{1}^{2-\alpha}\to +\infty (n\to\infty) \]
不科学。所以必然有$0<\alpha<2$,若$1<\alpha<2$,则
\[ \frac{a_{n}-a_{n+1}}{a_{n}^{2-\alpha}}\geq \beta \]
\[ \int_{a_{n+1}}^{a_{n}}\frac{1}{x^{2-\alpha}}dx\geq \beta \]
\[  +\infty>\int_{0}^{a_{1}}\frac{1}{x^{2-\alpha}}dx\geq \int_{a_{n+1}}^{a_{1}}\frac{1}{x^{2-\alpha}}dx\geq n\beta\]
同样不科学。所以必然有$0<\alpha\leq 1$.自然得到
\[ \int_{a_{n+1}}^{a_{n}}\frac{1}{x^{1-\alpha}}dx\geq \beta a_{n}\]
\[ +\infty>\int_{0}^{a_{1}}\frac{1}{x^{1-\alpha}}dx\geq \int_{a_{n+1}}^{a_{1}}\frac{1}{x^{1-\alpha}}dx\geq \beta S_{n}\]
知道$S_{n}$单调递增有上界,因此$\displaystyle\sum_{n=1}^{\infty}a_{n}<+\infty$,自然得到$a=0$.当$\alpha=1$时,
\[ (a_{n}-a_{n+1})\geq \beta a_{n}\qquad (n\geq N)\]
就是
\[ a_{N}\geq \beta\sum_{n=N}^{\infty}a_{n} \]
于是
\[ \sum_{n=N}^{\infty}a_{n}=O(a_{N}) \]
当$0<\alpha<1$时
\[ \frac{1}{\alpha}a_{N}^{\alpha}=\int_{0}^{a_{N}}\frac{1}{x^{1-\alpha}}dx\geq \beta\sum_{n=N}^{\infty}a_{n} \]
得到
\[ \sum_{n=N}^{\infty}a_{n}=O(a^{\alpha}_{N}) \]


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