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=(2)/(3)\\r=(1)/(2)`

The sum exists if
`|r|<1`

`\left|(1)/(2)\right|<1` therefore the sum exists

`\displaystyle\\\sum_(k=0)^(\infty)ar^k=(a)/(1-r)`

`(2)/(3)+(1)/(3)+(1)/(6)+\ldots=((2)/(3))/(1-(1)/(2))=((2)/(3))/((1)/(2))=(2)/(3)\cdot 2=(4)/(3)`