Calculating Infinite Sine Sum with Ratio Test | x and t Real Numbers

[Quacknetar]

Homework Statement

I have this exercise: Calculate
$\sum\limits_{k=0}^\infty t^{k}sin{(kx)}$
Where x and t are real and t is between 0 and 1.

Homework Equations

?

The Attempt at a Solution

The ratio test says that this sum does have a limit, and t^k obviously converges, as t is between 0 and 1. However, how am I supposed to solve it for the sines? A summatory inside a summatory? I don't think so, this is an exam question and we haven't done anything similar. I've been stuck with this for a day now...

 

[Orodruin]

How about trying to rewrite the sine in some intelligent fashion?

 

[Quacknetar]

Damn, how did I not think of that?
So the answer is
$\frac{2i}{2i-te^{ix}}-\frac{2i}{2i-te^{-ix}}$?

 

[Matternot]

I get:

$\frac{tsin(x)}{1-2t cos(x)+t^2}$

Your answer may be the same.

What was your method?

 

[Matternot]

$\frac{tsin(x)}{1-2tcos(x)+t^2}$

 

[LCKurtz]

Damn, how did I not think of that?
So the answer is
$\frac{2i}{2i-te^{ix}}-\frac{2i}{2i-te^{-ix}}$?

That isn't what I get. Show your work, and, since the answer must be real, put it in real form.

[Edit, added]I didn't see Stephen's post; I agree with his answer.

 

[Quacknetar]

Crap! I'm being stupid today! my previous expression should have been
$\frac{1}{2i}(\frac{1}{1-te^{ix}}-\frac{1}{1-te^{-ix}})$
Which, if simplified, gives the same answer you got. Thanks, guys!

 

[Matternot]

It seems like your method is unnecessarily complicated. Did you write sin as:

$\frac{e^{ix}-e^{-ix}}{2i}$

or as

$Im(e^{ix})$?

 

[Quacknetar]

I used
$\sin{(x)}=\frac{e^{ix}-e^{-ix}}{2i}$

How do you do it with

$Im(e^{ix})$?

 

[Orodruin]

I used
$\sin{(x)}=\frac{e^{ix}-e^{-ix}}{2i}$

How do you do it with

$Im(e^{ix})$?

In exactly the same way, you just move the I am outside of the sum since $t$ is real. The difference is that you get only one term.

 

[Matternot]

It's considered easier to use

$Im(e^{ix})$

Try using that method and see which you prefer.

 

[Quacknetar]

Okay, I've just tried it and it's definitely easier and there's less things that you can do wrong. I wouldn't have thought of that myself, as $Im(z)$ doesn't feel like a "proper" function to me, but I see it can be useful if you know what you're doing. Thanks for the help, guys!