Answer
$\dfrac{2}{5}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=0}^{\infty} \dfrac{2-3^k}{6^k}$ as $\Sigma_{k=0}^{\infty} [\dfrac{2}{6^k} - (\dfrac{3}{6})^{k}]$
or, $\Sigma_{k=0}^{\infty} [\dfrac{2}{6^k} - (\dfrac{3}{6})^{k}=3 \Sigma_{k=0}^{\infty} (\dfrac{1}{6} )^k-\Sigma_{k=0}^{\infty} (\dfrac{3}{6})^{k}$
Here, we can see that the series $3 \Sigma_{k=0}^{\infty} (\dfrac{1}{6} )^k-\Sigma_{k=0}^{\infty} (\dfrac{3}{6})^{k}$ shows a geometric series .
whose sum can be computed as: $\Sigma_{n=0}^{\infty} ar^n =a+ar+ar^2+ar^3+.....=\dfrac{a}{1-r}$
Now, $3 \Sigma_{k=0}^{\infty} (\dfrac{1}{6} )^k-\Sigma_{k=0}^{\infty} (\dfrac{3}{6})^{k}
=2 \times \dfrac{1}{1-1/6}- \dfrac{1}{1-3/6} \\= \dfrac{12}{5}-2\\=\dfrac{2}{5}$
