Answer
$\dfrac{113}{30}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=1}^{\infty} [\dfrac{1}{3} (\dfrac{5}{6})^{k}+\dfrac{3}{5}(\dfrac{7}{9})^{k}]$ as $\dfrac{1}{3} \Sigma_{k=1}^{\infty} (\dfrac{5}{6})^{k}+\dfrac{3}{5} \Sigma_{k=1}^{\infty} (\dfrac{7}{9})^{k}]$
Here, we can see that the series $\dfrac{1}{3} \Sigma_{k=1}^{\infty} (\dfrac{5}{6})^{k}+\dfrac{3}{5} \Sigma_{k=1}^{\infty} (\dfrac{7}{9})^{k}]$ shows a geometric series with common ratio $r=\dfrac{5}{6}\lt 1$ and $r=\dfrac{7}{9} \lt 1$. So, the series is convergent.
whose sum can be computed as: $\Sigma_{n=0}^{\infty} ar^n =a+ar+ar^2+ar^3+.....=\dfrac{a}{1-r}$
Now, $\dfrac{1}{3} \Sigma_{k=1}^{\infty} (\dfrac{5}{6})^{k}+\dfrac{3}{5} \Sigma_{k=1}^{\infty} (\dfrac{7}{9})^{k}]=\dfrac{1}{3} \times \dfrac{5/6}{1-\dfrac{5}{6}}+\dfrac{3}{5} \times \dfrac{7/9}{1-\dfrac{7}{9}} \\= \dfrac{5}{3}+ \dfrac{21}{10}\\=\dfrac{113}{30}$
