Question

2/3+(2/3)^2+(2/3)^3 + ... = x, find x.

Answer 1

Let S be the sum of the first n terms of the left side:

$S = \dfrac23 + \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \cdots + \left(\dfrac23\right)^n$

Multiply both sides by 2/3 :

$\dfrac23 S = \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \left(\dfrac23\right)^4 + \cdots + \left(\dfrac23\right)^{n+1}$

Subtract this from S :

$S - \dfrac23 S = \dfrac23 - \left(\dfrac23\right)^{n+1}$

Solve for S :

$\dfrac13 S = \dfrac23 - \left(\dfrac23\right)^{n+1}$

$S = 2 - 3 \left(\dfrac23\right)^{n+1}$

As n gets larger and larger, S converges to the given sum, and the term (2/3)ⁿ⁺¹ converges to zero, which leaves us with

$\displaystyle \lim_{n\to\infty} S = \boxed{x = 2}$