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

asked Feb 7, 2021 93.1k views

5 votes

by Asu

6.2k points

Please log in or register to add a comment.

2 votes

$a=(2)/(3)\\r=(1)/(2)$

The sum exists if
|r|<1

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

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

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

Aaronburrows answered Feb 14, 2021

5.9k points

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

5.8m questions

7.5m answers