The simplification of $$\frac{1}{8} + $$ $$\frac{1}{{{8^2}}} + $$ $$\frac{1}{{{8^3}}} + $$ $$\frac{1}{{{8^4}}} + $$ $$\frac{1}{{{8^5}}}$$ upto three place of decimals yields = ?

$$\eqalign{ & {\text{According to question,}} \cr & \frac{1}{8} + \frac{1}{{{8^2}}} + \frac{1}{{{8^3}}} + \frac{1}{{{8^4}}} + \frac{1}{{{8^5}}} \cr & \Rightarrow \frac{1}{8} + \frac{1}{{64}} + \frac{1}{{512}} + \frac{1}{{4096}} + \frac{1}{{32768}} \cr} $$
  ⇒ 0.125 + 0.015625 + 0.00195313 + 0.00024414 + 0.0000305175
  ⇒ 0.143