Question #242395

Sigma n=1 to infinity (-1)^n sin(1/n)

Expert's answer

This is an alternating series. We shall use alternating series test to check if it converges or not.

lim⁡n→∞sin⁡(1n)=sin⁡(0)=0\lim_{n\to \infty}\sin(\frac{1}{n})=\sin(0)=0

Hence, the series ∑n=1∞(−1)nsin⁡(1n)\sum_{n=1}^{\infty} (-1)^n \sin(\frac{1}{n}) is convergent by the alternating series test.


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