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$$\sum_{n\geq0} ({\frac{2}{3}})^{2n-1}$$

I managed to find that this is a geometric series with common ratio of $4/9$, therefore convergent. That means it's possible to calculate the sum of the series. Can I get any tip of how to do the sum of any kind of geometric series that converge? I've been researching, but all articles seem very confusing, specially with English, which I'm not native on.

Thanks!

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If this is homework, please consider tagging it as such. –  Aryabhata Nov 1 '10 at 18:10

3 Answers 3

up vote 3 down vote accepted

$$ (\frac{2}{3})^{2n-1} = \frac{3}{2}(\frac{4}{9})^n$$

If we have a geometric progression with common ratio $r$, and sum from 0 to infinity the sum is

$$s = \sum_{i=0}^\infty ar^i = \frac{a}{1-r}$$

if $|r|<1$ and where $a$ is the "common factor".

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$$S = 1 + x + x^2 + x^3 + \cdots, \quad |x|<1. \quad (1)$$ $$xS = x + x^2 + x^3 + \cdots \quad (2)$$ $$(1) - (2) \Rightarrow S-xS = 1.$$ $$S = \frac{1}{1-x}.$$

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No words, except these. –  Derek Jennings Nov 1 '10 at 18:13

$\displaystyle\rm\ \ \sum_{n\ge 0}\ r^{\:a+b\:n}\ =\ \frac{r^a}{1-r^b}\ \ if\ \ |r| < 1\ \ $ e.g. see the Wikipedia page on geometric series.

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For the sake of completeness you might add the case $|r|\ge 1$? –  AD. Nov 1 '10 at 18:55

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