Trigonometric series problem

I have the following problem from my Fourier analysis book I would need some guidance with. I have tried it, but apparently I made some mistakes...here is my problem:

We have:

$$\sin \theta = \frac{2}{\pi}-\frac{4}{\pi}\sum_1^{\infty}\frac{\cos 2n\theta}{4n^2-1}\;\;\;\;(0\leq\theta\leq \pi)\;\;\;\;(*),$$ and we also have:

$$\cos\theta=\frac{d}{d\theta}\sin\theta = -\int_{\pi/2}^\theta \sin\phi \:d\phi.$$

Show that series $(*)$ can be differentiated and integrated termwise to yield two apparently different expressions for $\cos\theta$ for $0 <\theta < \pi$, and reconcile these two expressions. (Hint the following equation is useful)

$$\sum_1^{\infty}\frac{(-1)^{n+1}}{n}\sin n\theta = \frac{\theta}{2}\;\;\text{for}\;\;-\pi < \theta < \pi.$$

Could someone give me some guidelines to this problem? =) I tried first differentiating $(*)$ and then integrating $(*)$, but I'm not sure if I got it right or not...

thnx for any guidance =)

My attempted solution:

$$\frac{d}{d\theta}\sin\theta = \frac{8}{\pi}\sum_1^{\infty}\frac{n}{4n^2-1}\sin2n\theta = \cos\theta$$

$$\cos\theta = -\int_{\pi/2}^\theta \sin\phi \;d\phi = -\int_{\pi/2}^\theta \left[\frac{2}{\pi}-\frac{4}{\pi}\sum_1^{\infty}\frac{\cos 2n\phi}{4n^2-1}\right]\;d\phi$$

$$= -\int_{\pi/2}^\theta \frac{2}{\pi}\;d\phi + \frac{4}{\pi}\int_{\pi/2}^\theta\sum_1^{\infty}\frac{\cos 2n\phi}{4n^2-1}\;d\phi = -\left[ \frac{2\phi}{\pi} \right]_{\pi/2}^{\theta} + \frac{4}{\pi}\left[ \sum_1^{\infty}\frac{\sin2n\phi}{2n(4n^2-1)}\right]_{\pi/2}^\theta$$

$$= -\left[\frac{2\theta}{\pi}-1\right] + \frac{4}{\pi}\left[ \sum_1^{\infty}\frac{\sin2n\theta}{2n(4n^2-1)}- \sum_1^{\infty}\frac{\sin n\pi}{2n(4n^2-1)} \right]$$

$$= 1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\frac{\sin2n\theta}{2n(4n^2-1)} = \cos\theta \;\;\;\text{?}$$

Here is where I get stuck...

So does:

$$1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\frac{\sin2n\theta}{2n(4n^2-1)} = \frac{8}{\pi}\sum_1^{\infty}\frac{n}{4n^2-1}\sin2n\theta = \cos\theta\;\;\;\;\;\;\;\; 0<\theta<\pi\;\;\;\;\text{?}$$

UPDATE :

It seems my calculations were correct, I tried visualizing my series in Matlab:

Now I just need to prove that the Fourier series are equal to each other =)

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And if you want to know whether you got it right or not then you should add your own work to your question... –  DonAntonio Mar 18 at 13:37
+1 Oh, sorry, it's typo. I'll correct it. Okay I will add my attempted solution. Actually, I didn't get the final solution...because my differentiation of $(*)$ and integration of $(*)$ were different...so...I dunno where I made my mistake.. –  jjepsuomi Mar 18 at 13:38
+1 Okay @DonAntonio will do :) Give me a sec to typo the latex ;) –  jjepsuomi Mar 18 at 13:43

Hints:

First of all:

$$\left|\frac{\cos 2n\theta}{4n^2-1}\right|\le\frac1{4n^2-1}\implies \;\text{since the series} \sum_{n=1}^\infty\frac1{4n^2-1}$$

converges, Weierstrass $\;M$-test gives so uniform convergence for all $\;\theta\in\Bbb R\;$ , which means we can differentiate-integrate termwise.

Differentiation:

$$(\sin\theta)'=\cos\theta\stackrel?=\left(\frac{2}{\pi}-\frac{4}{\pi}\sum_1^{\infty}\frac{\cos 2n\theta}{4n^2-1}\right)'=\frac4\pi\sum_{n=1}^\infty\frac{2n\sin2n\theta}{4n^2-1}$$

Integration:

$$\int\sin\theta d\theta=-\cos\theta\stackrel?=\int\left(\frac{2}{\pi}-\frac{4}{\pi}\sum_1^{\infty}\frac{\cos 2n\theta}{4n^2-1}\right)d\theta=$$

$$=\frac2\pi\theta-\frac4\pi\sum_{n=1}^\infty\frac{\sin2n\theta}{2n(4n^2-1)}$$

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+1 Thank you for your help @DonAntonio can you see my update =) I posted my attempt... –  jjepsuomi Mar 18 at 14:12
Sure @jjepsuomi, but now I've got to go. I'll try later. –  DonAntonio Mar 18 at 14:15
+1 No probs :) I understand, thank you! –  jjepsuomi Mar 18 at 14:16
Hi @DonAntonio did you have time yet to review my attempt? =) Is the direction correct? =) –  jjepsuomi Mar 20 at 7:14
I got it myself already :-) no need to answer :-) I will add the solution here soon :-) –  jjepsuomi Mar 20 at 12:44

I got the solution myself:

So the question was to prove that:

$$1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\frac{\sin2n\theta}{2n(4n^2-1)} = \frac{8}{\pi}\sum_1^{\infty}\frac{n}{4n^2-1}\sin2n\theta = \cos\theta\;\;\;\;\;\;\;\; 0<\theta<\pi\;\;\;\;$$

I start by using this equation:

$$1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\frac{\sin2n\theta}{2n(4n^2-1)} = \frac{8}{\pi}\sum_1^{\infty}\frac{n}{4n^2-1}\sin2n\theta\;\;\;\;\;\;\;\; 0<\theta<\pi$$

Moving the right hand side to left I get:

$$1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\frac{\sin2n\theta}{2n(4n^2-1)}-\frac{8}{\pi}\sum_1^{\infty}\frac{n}{4n^2-1}\sin2n\theta = 0$$

$$=1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\left(\frac{\sin2n\theta}{2n(4n^2-1)} - \frac{2n\sin 2n\theta}{4n^2-1}\right)$$

$$=1-\frac{2\theta}{\pi} + \frac{4}{\pi}\sum_1^{\infty}\frac{(4n^2-1)(-\sin2n\theta)}{2n(4n^2-1)} = 1-\frac{2\theta}{\pi} - \frac{4}{\pi}\sum_1^{\infty}\frac{\sin 2n\theta}{2n}=0$$

So I get:

$$1-\frac{2\theta}{\pi}=\frac{4}{\pi}\sum_1^{\infty}\frac{\sin 2n\theta}{2n}$$

or

$$\frac{\pi}{4}-\frac{\theta}{2}=\sum_1^{\infty}\frac{\sin 2n\theta}{2n}$$

It was given in the problem that:

$$\sum_1^{\infty}\frac{(-1)^{n+1}}{n}\sin n\theta = \frac{\theta}{2}\;\;\;\;\;-\pi<\theta<\pi$$

so

$$\frac{\pi}{4}-\sum_1^{\infty}\frac{(-1)^{n+1}}{n}\sin n\theta=\sum_1^{\infty}\frac{\sin 2n\theta}{2n},$$

which will result in after doing some algebra:

$$\frac{\pi}{4} = \sum_1^{\infty}\frac{\sin(2n-1)\theta}{2n-1}\;\;\;\;\;\;\;\; 0<\theta<\pi.$$

Now it is known that (from my book):

from which it follows:

$$1 = \frac{4}{\pi}\sum_1^{\infty}\frac{\sin(2n-1)\theta}{2n-1}\;\;\;\;\;\;\;\; 0<\theta<\pi.\;\;\;\;\blacksquare$$

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