# Evaluating an infinite series

## 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 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?

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Mark44
Mentor
What exactly are you supposed to do with this series?

What exactly are you supposed to do with this series?
Suppose to solve for the overall sum I'm assuming, the question simply states "Find $$\sum_{n=1}^{\infty}\frac{sinn}{2^n}$$."

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)
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.

There are two problems here:

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

(2) We can't simply change the lower index to n=1.
Well that was the only approach I could think of, any other method to solve this question? A starting tip/hint will suffice.

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.

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)
Calculus II (First year)

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.
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.

If you know Euler's formula (see http://en.wikipedia.org/wiki/Euler's_formula ), then you should try using that.
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

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)

Dick
Homework Helper
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?

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?
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?

Dick