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This is just a power series I came up with. I have plotted the series for up to $100$ terms and the plot seems to be pretty stable between over $(-1,1)$, but I'd like to verify that with a convergence test. At the least, I know that if the radius of convergence is $1$, then the interval is open since the series diverges for $x=\pm1$.

Neither the ratio nor root test seem to work, since the former yields a nonexistent limit and the latter gives $1$: $$\begin{align*}\textbf{Ratio test:}\quad\lim_{n\to\infty}\left|\frac{\sin(n+1)}{\sin n}\right|&=\lim_{n\to\infty}\left|\cos1+\sin1\cot n\right|\\[1ex]\textbf{Root test:}\quad\limsup_{n\to\infty}|\sin n|^{1/n}&\le\limsup_{n\to\infty}1^{1/n}=1\end{align*}$$ Is there another test I can use here?

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If $|x| < 1$, then the series converges absolutely:

$$|\sin(n) x^n | \le |x|^n$$

If $|x| > 1$, then: $\liminf \sin(n)x^n =-\infty$. The series fails the limit test.

Therefore $R =1$.

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To prove that $\sum_{k=1}^\infty\sin k$ does not converge, I will show that $\sin k$ does not converge to $0$ as $k\to\infty$. Suppose $\lim_{k\to\infty}\sin k=0$. Then, from $$ \sin(k+1)=\sin k\cos1+\cos k\sin1 $$ it would follow that also $\lim_{k\to\infty}\cos k=0$. But this is a contradiction, since $\sin^2k+\cos^2k=1$.

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