Lời giải chi tiết:
Ta có:
$\frac{m}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1334}+\frac{1}{1335}$
$=(1+\frac{1}{3}+...+\frac{1}{1333}+\frac{1}{1335})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1332}+\frac{1}{1334})$
$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1333}+\frac{1}{1334}+\frac{1}{1335})-2.(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1332}+\frac{1}{1334})$
$=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1333}+\frac{1}{1334}+\frac{1}{1335})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{666}+\frac{1}{667})$
$=\frac{1}{668}+\frac{1}{669}+...+\frac{1}{1334}+\frac{1}{1335}$