Assuming the coefficients are always 1, 4, 1, 4, 1, 4 so on:
\(\dfrac1{3^1} + \dfrac4{3^2} + \dfrac1{3^3} + \dfrac4{3^4} + \cdots\\ = \dfrac13 \left(1 + \dfrac19 + \dfrac1{9^2} + \cdots\right) + \dfrac4{3^2} \left(1 + \dfrac19 + \dfrac1{9^2}+\cdots\right)\\ = \left(\dfrac{1}{3} + \dfrac49\right)\left(\dfrac1{1 - \dfrac19}\right)\\ = \dfrac78\)