# Homework Help: Evaluating an infinite series

1. Mar 1, 2010

### Ali812

1. The problem statement, all variables and given/known data
$$\sum_{n=1}^{\infty}\frac{sinn}{2^n}$$

2. Relevant equations
Definition of a geometric series:
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

3. 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 the equation from part b and change the lower index to n=1)
$$\sum_{n=1}^{\infty}2^n=\frac{1}{1-2}$$

Taking the derivate of both sides:
$$\sum_{n=1}^{\infty}n2^{n-1}=\frac{1}{(1-2)^2}$$

Multiplying both sides by 2:
$$\sum_{n=1}^{\infty}n2^n=\frac{2}{(1-2)^2}$$

Of course I can simplify the question furthermore and get a single value as a result, my problem is I don't understand on how to incorporate the numerator (sin n) into the problem. Is my approach correct, or am I substituting the wrong x-value?

2. Mar 1, 2010

### Staff: Mentor

What exactly are you supposed to do with this series?

3. Mar 1, 2010

### Ali812

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

4. Mar 1, 2010

### Mathnerdmo

There are two problems here:

(1) The equation you gave for the geometric series only works for |x| < 1.

(2) We can't simply change the lower index to n=1.

5. Mar 1, 2010

### Ali812

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

6. Mar 1, 2010

### Mathnerdmo

I'm also having some difficulty with the sine in the numerator - I haven't done something like this in a while.

What class is this for, by the way? (to give me an idea of methods you have)

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Also, if there is a solution with geometric series, the problems I pointed out can be fixed... you just need to be careful about what you use.

7. Mar 1, 2010

### Ali812

Calculus II (First year)

8. Mar 1, 2010

### Ali812

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.

9. Mar 1, 2010

### Mathnerdmo

10. Mar 1, 2010

### Ali812

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

11. Mar 1, 2010

### Mathnerdmo

Hrmmm... I don't know a solution if you can't use Euler's formula. Sorry.

(doesn't mean there isn't one, but I'm not seeing it)

12. Mar 1, 2010

### Dick

I'm with Mathnerdmo. It's pretty easy if you put r=exp(i)/2 and realize your sum is related to the imaginary part of sum(r^n). So |r|<1 and you can treat it as a geometric series. Are you sure you didn't do that in Calc I and just forgot it?

13. Mar 2, 2010

### 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?

14. Mar 2, 2010

### Dick

exp(i)^n=exp(in)=cos(n)+i*sin(n). If you sum that series and take the imaginary part you get the sum of sin(n). What happens if you do the same thing with (exp(i)/2)^n?