[tex]\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{27}+\dfrac{8}{81}+\dfrac{16}{243} = \\ \\\\ = \sum\limits_{k=1}^{5}\dfrac{2^{k-1}}{3^k} = \sum\limits_{k=1}^{5}\dfrac{2^{k}}{2\cdot 3^k} = \dfrac{1}{2}\cdot \sum\limits_{k=1}^{5}\dfrac{2^{k}}{3^k} = \\ \\\\ = \dfrac{1}{2}\cdot \sum\limits_{k=1}^{5}\Big(\dfrac{2}{3}\Big)^k = \dfrac{1}{2}\cdot \left[\Big(\dfrac{2}{3}\Big)^1+\Big(\dfrac{2}{3}\Big)^2+...+\Big(\dfrac{2}{3}\Big)^5\right] =[/tex]
[tex]= \dfrac{1}{2}\cdot \dfrac{\dfrac{2}{3}\cdot\left[\Big(\dfrac{2}{3}\Big)^5-1\right]}{\dfrac{2}{3}-1} =-\Big(\dfrac{2}{3}\Big)^{5}+1 = \dfrac{-2^5+3^5}{3^5} = \boxed{\dfrac{211}{243}}[/tex]
[tex]\text{I used the geometric series formula for sum: }S_n = \dfrac{b_1\cdot (q^n - 1)}{q-1}[/tex]
