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How do I prove that 0.3333.. Is equal to 1/3 using a geometric series?

Sagot :

Answer:

Please see explanation.

Step-by-step explanation:

[tex]0.\overline{3}[/tex]

[tex]=\frac{1}{10}(3)+(\frac{1}{10})^2(3)+(\frac{1}{10})^3(3)+(\frac{1}{10})^4(3)\cdots[/tex]

[tex]=3 \cdot \frac{1}{10}(1+\frac{1}{10}+(\frac{1}{10})^2+(\frac{1}{10})^3 \cdots[/tex]

[tex]=3 \cdot \frac{1}{10}\sum_{i=1}^{\infty}(\frac{1}{10})^{n-1}[/tex]

[tex]=3 \cdot \frac{1}{10} \cdot \frac{1}{1-\frac{1}{10}}[/tex]

[tex]=\frac{3}{10} \cdot \frac{10}{10-1}[/tex]

[tex]=\frac{3}{10} \cdot \frac{10}{9}[/tex]

[tex]=\frac{3}{9}[/tex]

[tex]=\frac{1}{3}[/tex]

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