# Infinite sum of sines

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1. Apr 8, 2015

### Quacknetar

1. The problem statement, all variables and given/known data
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.
2. Relevant equations
?
3. The attempt at a solution
The ratio test says that this sum does have a limit, and tk 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...

Last edited: Apr 8, 2015
2. Apr 8, 2015

### Orodruin

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

3. Apr 8, 2015

### Quacknetar

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

4. Apr 8, 2015

### Matternot

I get:

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

Last edited: Apr 8, 2015
5. Apr 8, 2015

### Matternot

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

Last edited: Apr 8, 2015
6. Apr 8, 2015

### LCKurtz

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

7. Apr 8, 2015

### 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!

8. Apr 9, 2015

### Matternot

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

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

or as

$Im(e^{ix})$?

9. Apr 9, 2015

### Quacknetar

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

How do you do it with

$Im(e^{ix})$?

10. Apr 9, 2015

### Orodruin

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

11. Apr 9, 2015

### Matternot

It's considered easier to use

$Im(e^{ix})$

Try using that method and see which you prefer.

12. Apr 9, 2015

### 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!