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So, $\sum_{n=2}^\infty|\sin{n}|^n$ converges. |
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Since $\pi$ is transcendental (so also $\frac{\pi}{2}$ and $\frac{3\pi}{2}$), $\forall n>0 n \in \mathbb{N} , |\sin{n}|<1$. In another hand, $\sum_{n=2}^\infty|\sin{n}|^n <\sum_{n=2}^\infty|\sin{n}|^2$ which converges (because $\sum_{n=1}^\infty a^n$ converges if $|a| < 1$. So, $\sum_{n=2}^\infty|\sin{n}|^n$ converges. |
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Since $\pi$ is transcendental (so also $\frac{\pi}{2}$ and $\frac{3\pi}{2}$), $\forall n>0 , |\sin{n}|<1$. In another hand, $\sum_{n=2}^\infty|\sin{n}|^n <\sum_{n=2}^\infty|\sin{n}|^2$ which converges (because $\sum_{n=1}^\infty a^n$ converges if $a |a| < 1$. So, $\sum_{n=2}^\infty|\sin{n}|^n$ converges. |
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