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

Sagot :

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

[tex]S = \dfrac23 + \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \cdots + \left(\dfrac23\right)^n[/tex]

Multiply both sides by 2/3 :

[tex]\dfrac23 S = \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \left(\dfrac23\right)^4 + \cdots + \left(\dfrac23\right)^{n+1}[/tex]

Subtract this from S :

[tex]S - \dfrac23 S = \dfrac23 - \left(\dfrac23\right)^{n+1}[/tex]

Solve for S :

[tex]\dfrac13 S = \dfrac23 - \left(\dfrac23\right)^{n+1}[/tex]

[tex]S = 2 - 3 \left(\dfrac23\right)^{n+1}[/tex]

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

[tex]\displaystyle \lim_{n\to\infty} S = \boxed{x = 2}[/tex]