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Alternating series

S1=n=1(1)nsin(n+1)S2=n=1sin(n+1)\begin{aligned} &S_1=\displaystyle\sum\limits_{n=1}^{\infty }{(-1)^n \sin(n+1)} \\\\ &S_2=\displaystyle\sum\limits_{n=1}^{\infty }{\sin(n+1)} \end{aligned}
Which series converges?
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Based on your answer, what can we say about S1S_1?
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