Answer
$\dfrac{3313}{3300}$
Work Step by Step
We are given the repeating decimal: $1.00\overline{39}=1.00393939....$
Rewrite the given repeating decimal:
$1.00393939....=1.0039+0.000039+0.00000039+.......$
$=1+0.0039+0.000039+0.00000039+.......$
$=1+\sum_{k=0}^{\infty} 0.0039\cdot 10^{-2k}$
The infinite geometric series $\sum_{k=0}^{\infty} a_1r^k$ has the sum $S=\dfrac{a_1}{1-r}$.
Determine the value of the repeating decimal:
$1+\sum_{k=0}^{\infty} 0.0039\cdot 10^{-2k}=1+\dfrac{0.0039}{1-0.01}=1+\dfrac{\dfrac{39}{10,000}}{\dfrac{99}{100}}=1+\dfrac{39}{9900}=1+\dfrac{13}{3300}=\dfrac{3313}{3300}$
