Express the infinite series $$\frac{3}{206}+\frac{9}{2\cdot103^2}+\frac{27}{2\cdot103^3}+\cdots$$as a terminating decimal.
I notice that this series has a geometric structure: each term is obtained by multiplying the previous term by $\frac{3}{103}$. So, I can use the formula for the sum of an infinite geometric series: $$S = \frac{a}{1-r}$$ where $a$ is the first term and $r$ is the common ratio. In this case, $a = \frac{3}{206}$ and $r = \frac{3}{103}$. So, the sum is $$S = \frac{\frac{3}{206}}{1-\frac{3}{103}} = \frac{3}{206-6} = \frac{3}{200} = 0.015$$ 