Limit Problem: Solving sin(kθ)/2k for n→∞

[MATHMAN89]

Homework Statement

The problem is as follows:

lim(n→∞)(sum(sin(kθ)/2^k,k=1,n)

Homework Equations

The Attempt at a Solution

I feel like this should converge since sin oscillates between -1 and 1 and 2^k keeps getting larger and larger.

Do I have to use power series somehow?

Any ideas, anyone?

 

[Mark44]

Homework Statement

The problem is as follows:

lim(n→∞)(sum(sin(kθ)/2^k,k=1,n)

Homework Equations

The Attempt at a Solution

I feel like this should converge since sin oscillates between -1 and 1 and 2^k keeps getting larger and larger.

Do I have to use power series somehow?

Any ideas, anyone?

Are you trying to determine whether this series converges?
$$\sum_{k = 1}^\infty \frac{sin(k\theta)}{2^k}$$

What tests do you know?

 

[MATHMAN89]

Are you trying to determine whether this series converges?
$$\sum_{k = 1}^\infty \frac{sin(k\theta)}{2^k}$$

What tests do you know?

Thanks for your response.

The problem can be rewritten as sum (a_k*(1/2)^k) where a_k=sin(kθ)
Then I can use limsup ( ⁿ√abs(a_k) = 1
Therefore my radius of convergence is 1. But now how do I use that to find the limit?

 

[Dick]

Your series is related to the geometric series a^k where a=e^(i*theta)/2. How is it related?

 

[MATHMAN89]

Your series is related to the geometric series a^k where a=e^(i*theta)/2. How is it related?

I see what you mean. If I do ((e^i*theta)/2)^k I would get

(cos(kθ)+isin(kθ))/2^k
I just need to get the numerator equal to sin(kθ)
How do I do that?? I need to get rid of the cos (kθ) and the i in front of sin

Could I do(e^i*theta - e^-i*theta / 2i)^k = sin(k*theta) ?

 

[Dick]

I see what you mean. If I do ((e^i*theta)/2)^k I would get

(cos(kθ)+isin(kθ))/2^k
I just need to get the numerator equal to sin(kθ)
How do I do that?? I need to get rid of the cos (kθ) and the i in front of sin

Could I do(e^i*theta - e^-i*theta / 2i)^k = sin(k*theta) ?

No, you don't want to do that. Just take the imaginary part of the complex series.

 

[MATHMAN89]

No, you don't want to do that. Just take the imaginary part of the complex series.

Aha. And then I have

lim n →∞ of I am ( sum (e^iθ/2)^k , k = 1 to n)
So this is a geometric series and would be equal to
lim n →∞ of I am ( e^iθ/2 * (1-((e^iθ)/2)^n)/(1-(e^iθ)/2)

 

[MATHMAN89]

I got sin(theta)/(2-sin(theta)) as my answer. Sound right?

 

[Dick]

I got sin(theta)/(2-sin(theta)) as my answer. Sound right?

Nope. I don't think you were careful enough separating real and imaginary. Try and think of a way to check it. Suppose you put theta=pi/2 in your series. Can you sum it? Then put pi/2 in your formula.