头条数学救火队长马丁的一道中山大学研究生入学考试数学分析题

头条数学救火队长马丁老师的题目，
中山大学考研数学分析试题

\(假设级数\sum_{i=1}^{n}发散，且\{a_{n}\}是单调不增非负数列，试证：\)
\(\quad\quad\quad \lim_{n\to\infty}\frac{a_{2}+a_{4}+a_{6}\cdot\cdot\cdot a_{2n}}{a_{1}+a_{3}+a_{5}\cdot\cdot\cdot a_{2n-1}}=1\)
\(\\\)
\(证明\)
\(\quad\lvert \frac{a_{2}+a_{4}+a_{6}\cdot\cdot\cdot a_{2n}}{a_{1}+a_{3}+a_{5}\cdot\cdot\cdot a_{2n-1}}-1 \rvert\\\)
\(=\lvert \frac{a_{2}-a_{1}+a_{4}-a_{3}+a_{6}-a_{5}+\cdot\cdot\cdot+a_{2n}\quad-\quad_{2n-1}}{a_{1}+a_{3}+a_{5}\cdot\cdot\cdot a_{2n-1}} \quad\rvert\\\)
\(=\frac{a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+\cdot\cdot\cdot + a_{2n-1}\quad-\quad_{2n}}{a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot +a_{2n-1}}\\\)
\(\leqslant\frac{a_{1}-a_{2}+a_{2}-a_{4}+a_{4}-a_{6}+\cdot\cdot\cdot +a_{2n-2}\quad-\quad a_{2n}}{a_{1}+a_{3}+a_{5}\cdot\cdot\cdot +a_{2n-1}}\\\)
\(\leqslant\frac{a_{1}-a_{2n}}{a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot+ a_{2n-1}}\\\)
\(\leqslant\frac{a_{1}}{a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot+ a_{2n-1}}\quad\quad\quad(1)\\\)
\(\\\)
\(若\lim_{n\to\infty}a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot +a_{2n-1}=A\)
\(则\)
\(lim_{n\to\infty}a_{1}+a_{2}+a_{3}++a_{4}+\cdot\cdot\cdot +a_{2n-1}+a_{2n}\)
\(\leqslant\lim_{n\to\infty}a_{1}+a_{1}+a_{3}+a_{3}+\cdot\cdot\cdot +a_{2n-1}+a_{2n-1}\)
\(=2A\)
与题设矛盾
\(故\lim_{n\to\infty}a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot+ a_{2n-1}=+\infty\)
\(即,\forall N\in N^+,\exists N_{0},当n>N_{0},有a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot+ a_{2n-1}>N\)
\(故\forall \epsilon >0,\exists N 当n>N时，有a_{1}+a_{3}+a_{5}+\cdot\cdot\cdot+ a_{2n-1}>\frac{a_{1}}{\epsilon}\)
\(即\quad \frac{a_{1}}{a_{1}+a_{2}+\cdot\cdot\cdot+a_{2n-1}}<\epsilon\)
\(即,当n>N时，（1）式<\epsilon\)
\(证毕\)