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$$\sum_{n=1}^{\infty}\frac{1}{2^n3^{(n-1)}}$$

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    Note that you have a geometric series whose terms are decreasing.2010-11-30
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    @Mohammed Alnasiri: Please check that I got the edit right. Thanks2010-11-30

2 Answers 2

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Your series can be written as $$\sum_{n=1}^\infty \frac {3}{6^n}.$$ It is a geometric series with $a=3/6=1/2, r=1/6$.

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You could use the comparison test:

\[\sum_{n=1}^\infty \frac{1}{2^n 3^{n-1}}<\sum_{n=1}^\infty \frac{1}{2^n}=1.\]