Sum of (-1)^(k+1) * sin (kx)/k

[robertoCamera]

Homework Statement

In short, I want to find \sum^{\infty}_{k=1} (-1)^{k+1} sin(kx)/k
I happen to know(/strongly believe) this is equal to x/2, but I'm not sure how to show it.

If we could show that sum k=1 to infinity of (-e^2pi i x/p )^k=-e^(2pi i x/p)/(1+e^(2pi i x/p)) that would suffice ( I believe), because the derivative would match.

see this post:
https://www.physicsforums.com/showthread.php?p=2923108#post2923108

Homework Equations

Letting f(x)=x defined on [-p/2,p/2), then extended periodically to the reals, the Fourier series of f is g(x)=(p/\pi)\sum_{k \neq 0} (-1)^{k+1} e^{2\pi i kx/p} p/(2\pi i k)=\sum^{\infty}_{k=1} (-1)^{k+1} sin(2\pi kx/p)/k

The Attempt at a Solution

I have already shown that the series converges uniformly on (-p/2,p/2), but I have not yet shown that it converges uniformly to x.

I did this by showing |S_{n+p} -S_n |^2 \leq (p^2/k^2) (\sum^{n+p}_{k=n+1} sin^2(kx))(\sum^{n+p}_{k=n+1}(1/k^2) which converges to 0. This means that the series converges uniformly by the Cauchy Criterion.

As for actually evaluating \sum^{\infty}_{k=1} (-1)^{k+1} sin(2\pi kx/p)/k I don't really know where to begin.

Thank you for any help.BTW how does on embed Tex on this forum?

 

[mighty2000]

Hi there,

To embed Tex on this forum, you can use the following format:

$$insert your Tex code here$$

For example, if you want to write the expression "x^2 + y^2 = z^2", you can use the following code:

$$x^2 + y^2 = z^2$$

This will display as: x^2 + y^2 = z^2

Now, onto your question - finding the value of \sum^{\infty}_{k=1} (-1)^{k+1} sin(kx)/k. It seems like you have made some good progress in showing that the series converges uniformly. To actually evaluate the series, one approach would be to use the fact that the series is the Fourier series of the function f(x) = x on the interval [-p/2, p/2). This means that the value of the series at any point x will be equal to the value of f(x) at that point. In other words, the series converges to the function f(x) = x.

To prove this, you can use the fact that the Fourier series of f(x) is given by g(x) = (p/pi) * \sum_{k \neq 0} (-1)^{k+1} e^{2\pi i kx/p}. You can show that this series is equal to the series you are trying to evaluate, and then use the fact that the Fourier series of a function is equal to the function itself to show that the series converges to f(x) = x.

I hope this helps. Let me know if you have any other questions. Good luck!