Can You Prove the Convergence of a Trigonometric Series?

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SUMMARY

The convergence of the trigonometric series \(\sum\limits_{n=0}^\infty \frac{\sin^{4}(\frac{n\pi}{4})}{n^2}\) can be proven to equal \(\frac{\pi^{2}}{16}\). The key insight involves recognizing the periodic nature of \(\sin(\frac{n\pi}{4})\), which takes on a limited set of values. By leveraging the known result \(\sum\limits_{n=0}^\infty \frac{1}{n^2} = \frac{\pi^{2}}{8}\) for odd \(n\), one can effectively split the series into manageable parts to facilitate the proof.

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Homework Statement



I need to show that \sum\limits_{n=0}^\infty \frac{sin^{4}(\frac{n\pi}{4})}{n^2} = \frac{\pi^{2}}{16}

Homework Equations



I have this property for odd n

\sum\limits_{n=0}^\infty \frac{1}{n^2} = \frac{\pi^{2}}{8}

The Attempt at a Solution


[/B]
I have no idea how to do this, any help?
 
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There are only so many values that ##\sin(\frac{n\pi}{4})## can take. How about splitting the sum up on that basis?
 

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