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：)
(quadquadquad lim_{n oinfty}frac{a_{2}+a_{4}+a_{6}cdotcdotcdot a_{2n}}{a_{1}+a_{3}+a_{5}cdotcdotcdot a_{2n-1}}=1)
(\)
(证明)
(quadlvert frac{a_{2}+a_{4}+a_{6}cdotcdotcdot a_{2n}}{a_{1}+a_{3}+a_{5}cdotcdotcdot a_{2n-1}}-1
vert\)
(=lvert frac{a_{2}-a_{1}+a_{4}-a_{3}+a_{6}-a_{5}+cdotcdotcdot+a_{2n}quad-quad_{2n-1}}{a_{1}+a_{3}+a_{5}cdotcdotcdot a_{2n-1}} quad
vert\)
(=frac{a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+cdotcdotcdot + a_{2n-1}quad-quad_{2n}}{a_{1}+a_{3}+a_{5}+cdotcdotcdot +a_{2n-1}}\)
(leqslantfrac{a_{1}-a_{2}+a_{2}-a_{4}+a_{4}-a_{6}+cdotcdotcdot +a_{2n-2}quad-quad a_{2n}}{a_{1}+a_{3}+a_{5}cdotcdotcdot +a_{2n-1}}\)
(leqslantfrac{a_{1}-a_{2n}}{a_{1}+a_{3}+a_{5}+cdotcdotcdot+ a_{2n-1}}\)
(leqslantfrac{a_{1}}{a_{1}+a_{3}+a_{5}+cdotcdotcdot+ a_{2n-1}}quadquadquad(1)\)
(\)
(若lim_{n oinfty}a_{1}+a_{3}+a_{5}+cdotcdotcdot +a_{2n-1}=A)
(则)
(lim_{n oinfty}a_{1}+a_{2}+a_{3}++a_{4}+cdotcdotcdot +a_{2n-1}+a_{2n})
(leqslantlim_{n oinfty}a_{1}+a_{1}+a_{3}+a_{3}+cdotcdotcdot +a_{2n-1}+a_{2n-1})
(=2A)
与题设矛盾
(故lim_{n oinfty}a_{1}+a_{3}+a_{5}+cdotcdotcdot+ a_{2n-1}=+infty)
(即,forall Nin N^+,exists N_{0},当n>N_{0},有a_{1}+a_{3}+a_{5}+cdotcdotcdot+ a_{2n-1}>N)
(故forall epsilon >0,exists N 当n>N时，有a_{1}+a_{3}+a_{5}+cdotcdotcdot+ a_{2n-1}>frac{a_{1}}{epsilon})
(即quad frac{a_{1}}{a_{1}+a_{2}+cdotcdotcdot+a_{2n-1}}<epsilon)
(即,当n>N时，（1）式<epsilon)
(证毕)