$$\sum_{n=1}^{\infty}\frac{1}{2^n3^{(n-1)}}$$
whether the following series converge or diverge
0
$\begingroup$
calculus
-
3Note that you have a geometric series whose terms are decreasing. – 2010-11-30
-
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 $a=3/6=1/2, r=1/6$.
2
You could use the comparison test:
\[\sum_{n=1}^\infty \frac{1}{2^n 3^{n-1}}<\sum_{n=1}^\infty \frac{1}{2^n}=1.\]