Question

Evaluate 2/3 + 1/3 + 1/6 + … THIS IS CONTINUOUS. It is NOT as simple as 2/3 + 1/3 + 1/6.

Answer 1

$a=\dfrac{2}{3}\\r=\dfrac{1}{2}$

The sum exists if $|r|$

$\left|\dfrac{1}{2}\right|$ therefore the sum exists

$\displaystyle\\\sum_{k=0}^{\infty}ar^k=\dfrac{a}{1-r}$

$\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\ldots=\dfrac{\dfrac{2}{3}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{2}}=\dfrac{2}{3}\cdot 2=\dfrac{4}{3}$