\[\boxed{\mathbf{908\ (908).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[Пусть\ прогрессия\ b_{1},\ b_{2},\ b_{3},\ \]
\[\ldots,b_{n},\ldots;\ \ |q| < 1.\]
\[S = \frac{b_{1}}{1 - q}:\]
\[\frac{S}{S - b_{1}} = \frac{b_{1}}{1 - q}\ :\left( \frac{b_{1}}{1 - q} - b_{1} \right) =\]
\[= \frac{b_{1}}{1 - q}\ :\left( \frac{b_{1} - b_{1}(1 - q)}{1 - q} \right) =\]
\[= \frac{b_{1}}{1 - q}\ :\frac{b_{1} - b_{1} + b_{1}q}{1 - q} =\]
\[= \frac{b_{1}}{1 - q} \cdot \frac{1 - q}{b_{1}q} = \frac{b_{1}}{b_{1}q} = \frac{b_{1}}{b_{2}}\]
\[\frac{b_{1}}{b_{2}} = \frac{b_{1}}{b_{2}} \Longrightarrow доказано.\]