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# Evaluate the infinite geometric series: ​

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Evaluate the infinite geometric series: $$\frac13+\frac16+\frac{1}{12}+\frac{1}{24} \dots$$

May 2, 2022

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First term is 1/3.

Common ratio is (1/6)/(1/3) = 1/2.

Using the formula sum = a/(1 - r), sum = $$\dfrac{\frac13}{1 - \frac12} = \dfrac23$$

May 2, 2022
edited by MaxWong  May 2, 2022
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Thanks so much!

@MaxWong and @Anthrax

ItsFree  May 2, 2022
edited by ItsFree  May 2, 2022
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Observe that we can say

$$\displaystyle\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \cdots = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{4}+\cdots\right) = \frac{1}{3}\sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n.$$

Given this, our common ratio $$r$$ is $$r = \frac{1}{2}$$. Then, by the well-known infinite geometric series formula, we have

$$\displaystyle \frac{1}{3} \sum^{\infty}_{n=0}\left(\frac{1}{2}\right)^n = \frac{1}{3} \cdot\frac{1}{1-\frac{1}{2}} = \frac{2}{3}.$$

May 2, 2022