Series - convergent or divergent?

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