jmc

We start by factoring $\frac{1}{2}$ from all terms in the series to get $$\frac{1}{2}\left(\frac{3}{103}+\frac{9}{103^2}+\frac{27}{103^3}+\cdots \right).$$Next we recognize the series as a geometric series and apply the formula for the sum of a geometric series $\left(\frac{a}{1-r}\right)$: $$\frac{1}{2}\left(\frac{\frac{3}{103}}{1-\frac{3}{103}}\right)=\frac{1}{2}\left(\frac{3}{103-3}\right)=\frac{1}{2}\cdot \frac{3}{100}.$$To turn the fraction into a terminating decimal, we recognize that $\frac{3}{100}=0.03$ and half of 0.03 is $\boxed{0.015}$.