$\sum_{n=1}^{\infty}\frac{1}{2^n3^{(n-1)}}$
whether the following series converge or diverge
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calculus
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0@Mohammed Alnasiri: Please check that I got the edit right. Thanks – 2010-11-30
2 Answers
5
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:
\[\sum_{n=1}^\infty \frac{1}{2^n 3^{n-1}}<\sum_{n=1}^\infty \frac{1}{2^n}=1.\]