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How can I check if the following sum converges/diverges? I don't believe the comparison test will give me the result in this specific case. $\sum_{k=1}^{\infty}\sin (\frac{k^{2}+1}{k}\pi )$

1 Answers1

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Hint: $\sin (\frac{k^{2}+1}{k}\pi)=\sin (k+\frac{1}{k})\pi=\cos k\pi \sin\frac{\pi}{k}=(-1)^k\sin\frac{\pi}{k}$
Now use alternating series test.

Vasili
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