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 = \frac23 + \left(\frac23\right)^2 + \left(\frac23\right)^3 + \cdots + \left(\frac23\right)^n`

Multiply both sides by 2/3 :

`\frac23 S = \left(\frac23\right)^2 + \left(\frac23\right)^3 + \left(\frac23\right)^4 + \cdots + \left(\frac23\right)^(n+1)`

Subtract this from S :

`S - \frac23 S = \frac23 - \left(\frac23\right)^(n+1)`

Solve for S :

`\frac13 S = \frac23 - \left(\frac23\right)^(n+1)`

`S = 2 - 3 \left(\frac23\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}`