Verify that a sum converges to particular function (Fourier Series)

GamesMasta
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Homework Statement


Verify the formula x=2*(sin(x)-(1/2)sin(2x)+(1/3)sin(3x)-...), {x,-Pi,Pi}

Homework Equations


The Attempt at a Solution


I guess, I am to show that the sum on the right converges to the function x. I began by rewriting the sum on the RHS as $\displaystyle\sum_{k=1}^k 2*\frac{1}{k}*sin(kx)(-1)^{k-1}$

Now, I'm not sure what I am to do next. Am I to take the limit of k to infinity? If so how does one solve that? Thank you!

Also, if one graphs this in Mathematica, it can be seen that as k becomes larger and larger the sin function becomes more and more like x through the origin between -Pi and Pi. I believe, however, that I am to show this algebraicly...
 
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Do you know how to calculate Fourier series coefficients? I assume all you're supposed to do is verify that they match what you're given
 
Office_Shredder said:
Do you know how to calculate Fourier series coefficients? I assume all you're supposed to do is verify that they match what you're given

Ah, yes I do! I was assuming that since the sum was written that was supposed to use that. I guess if I attempt to calculate the function with only sine terms the coefficients would be given by: b_{k}=\frac{2}{Pi}\int^{Pi}_{0}f(x)sin(kx)dx. You obtain (-1)^(k-1)(2/k) for the coefficient. This then gives you the Fourier Series that gives you the RHS of the original equation. But I am now to prove convergence? That is something I am having difficulty understanding. Any help would be appreciated.
 
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Bump.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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