Verify Infinite Series: Proving Cosine Sum and Integrals

jj1986
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Homework Statement



Show that for all integers n \geq 1,
cos(2x) + cos(4x) + ... + cos(2nx) = \frac{1}{2} (\frac{sin((2n+1)x)}{sin(x)}-1)

Use this to verify that
\sum_{n=1}^{\infty}(\int_{0}^{\pi} x(\pi-x)cos(2nx)dx) =

\frac{-1}{2}\int_{0}^{\pi} x(\pi-x)dx)


Homework Equations





The Attempt at a Solution


I proved the first part of this problem using induction, however I don't see how I can use that to verify the second part. Maybe I can bring the summation into the integral and get the sum of cos(2nx), but I still don't see how that would give me what I need to prove. Any suggestions?
 
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If you interchange the sum with the integral, you can substitute the identity you have proved.
Then the second part (with the - 1) will give you the answer you want, you will just need to prove that
\lim_{n \to \infty} \frac12 \int_0^\pi x (\pi - x) \frac{\sin((2n+1)x)}{\sin(x)}\, dx = 0
 
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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