Does the series converge? (You can only use Ratio Test)
[math]\sum_{n=1}^\infty n^2\sin\frac{\pi}{2^n},\,\,\,\,a_n=n^2\sin\frac{\pi}{2^n} \,\,\,\,\Rightarrow\,\,\,\,a_{n+1}=(n+1)^2\sin\frac{\pi}{2^{n+1}}\\\\\frac{a_{n+1}}{a_n}=\left(\frac{n+1}{n}\right)^2\cdot\frac{\sin\frac{\pi}{2^{n+1}}}{\sin\frac{\pi}{2^n}}=\left(\frac{n+1}{n}\right)^2\frac{\sin\frac{\pi}{2^{n+1}}}{2\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}}=\left(\frac{n+1}{n}\right)^2\frac{1}{2\cos\frac{\pi}{2^{n+1}}}\\\\\text{and I'm stuck here guys, do you have any propositions what is next? or} \\\text{my idea to use sin double angle formula is a wrong way in that case...}[/math]
[math]\sum_{n=1}^\infty n^2\sin\frac{\pi}{2^n},\,\,\,\,a_n=n^2\sin\frac{\pi}{2^n} \,\,\,\,\Rightarrow\,\,\,\,a_{n+1}=(n+1)^2\sin\frac{\pi}{2^{n+1}}\\\\\frac{a_{n+1}}{a_n}=\left(\frac{n+1}{n}\right)^2\cdot\frac{\sin\frac{\pi}{2^{n+1}}}{\sin\frac{\pi}{2^n}}=\left(\frac{n+1}{n}\right)^2\frac{\sin\frac{\pi}{2^{n+1}}}{2\sin\frac{\pi}{2^{n+1}}\cos\frac{\pi}{2^{n+1}}}=\left(\frac{n+1}{n}\right)^2\frac{1}{2\cos\frac{\pi}{2^{n+1}}}\\\\\text{and I'm stuck here guys, do you have any propositions what is next? or} \\\text{my idea to use sin double angle formula is a wrong way in that case...}[/math]
