Recent content by Ali812

After skimming through some info on Euler's formula, I can substitute sinn= (e^in - e^(-in)) / 2i, I'm confused as to what you are trying to state, could you perhaps elaborate? How exactly did you derive with r = exp(i)/2?

I believe we haven't learned that yet unfortunately, however I have determined that the series is absolutely convergent via the comparison test. What r value can i use so |r| < 1 , involved with the numerator as well? :S

From what I can tell, you can change the lower index to n=1, but I assume I will have to check for convergence first, if the series does converge, I can move ahead with this method, if not, I will need an alternative.

Calculus II (First year)

Well that was the only approach I could think of, any other method to solve this question? A starting tip/hint will suffice.

Suppose to solve for the overall sum I'm assuming, the question simply states "Find \sum_{n=1}^{\infty}\frac{sinn}{2^n} ."

Homework Statement \sum_{n=1}^{\infty}\frac{sinn}{2^n} Homework Equations Definition of a geometric series: \sum_{n=0}^{\infty}x^n=\frac{1}{1-x} The Attempt at a Solution Basically I can use the geometric series idea and implement it into the denominator of the question (i.e. sub x=2 into...