This is an alternating series. We shall use alternating series test to check if it converges or not.
limn→∞sin(1n)=sin(0)=0\lim_{n\to \infty}\sin(\frac{1}{n})=\sin(0)=0limn→∞sin(n1)=sin(0)=0
Hence, the series ∑n=1∞(−1)nsin(1n)\sum_{n=1}^{\infty} (-1)^n \sin(\frac{1}{n})∑n=1∞(−1)nsin(n1) is convergent by the alternating series test.
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