[tex]a=\dfrac{2}{3}\\r=\dfrac{1}{2}[/tex]
The sum exists if [tex]|r|<1[/tex]
[tex]\left|\dfrac{1}{2}\right|<1[/tex] therefore the sum exists
[tex]\displaystyle\\\sum_{k=0}^{\infty}ar^k=\dfrac{a}{1-r}[/tex]
[tex]\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\ldots=\dfrac{\dfrac{2}{3}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{2}}=\dfrac{2}{3}\cdot 2=\dfrac{4}{3}[/tex]
