# Series - convergent or divergent?

$\displaystyle \sqrt{2}=2cos\frac{\pi}{4}$, we see that $\displaystyle a_n=\sqrt{2-2cos\frac{\pi}{2^n}}=\sqrt{4sin^2\frac{\pi}{2^{n+1 }}}=2sin\frac{\pi}{2^{n+1}}$. then $\displaystyle \sum\limits^{\infty}_{n=1}\frac{\pi}{2^n}$ converges and $\displaystyle 2sin\frac{\pi}{2^{n+1}}<\frac{\pi}{2^n}$ so your series is convergent
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