Find the sum of the series.

Question

anonymous · Answer

$$\sum_{n=1}^{\infty} (-3)^{n-1}/2^{3n}$$

anonymous · Answer

i am thinking. 
i got 1/3 but wolfram is telling me it is 1/11 so i must be doing something wrong
idea is to sum a geometric series using 
$$\frac{a}{1-r}$$

anonymous · Answer

= - 1/3 ( - 3/8) / (11/8 )

anonymous · Answer

that works for geometric series where a= a constant number but since -3 is raised to the n-1, it will keep oscillating between -3 and 3 and therefore it is not constant so you can not use that property.

anonymous · Answer

ok i see my mistake
first term is 
$$\frac{1}{8}$$ common ratio is 
$$-\frac{3}{8}$$ so should be 
$$\frac{\frac{1}{8}}{1+\frac{3}{8}}=\frac{1}{11}$$

anonymous · Answer

no it is not going from -3 to 3

anonymous · Answer

oh I see!! That makes more sense!! Sorry about that.

anonymous · Answer

it is 
$$(-3)^{n-1}$$