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If I know that $\sin(x)/x \geq 2/\pi$ for $x$ in $[0, \pi/2]$, how can I deduce that $\sum_1^\infty \sin(1/n)$ diverges? I'm not seeing any connection between the two. By inspection, it looks like the series converges to 1, but I don't even need to know the limit. I just need to argue that it converges.

My professor has been comparing series and sequences. I have used the harmonic series to show others diverge. I've used Euler's series of inverse squares to show others converge, so there's probably a comparison here that can do this one as well.

user1145880
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    Actually $\sin(x)/x \geq 2/\pi$ for $0 < x \le \pi/2$, and you can use that to show that $\sum_1^\infty \sin(1/n)$ diverges. – Martin R Mar 29 '24 at 16:09
  • Thanks, @MartinR. I had just noticed the inequality mistake, but I did think it the series converges. – user1145880 Mar 29 '24 at 16:12
  • @MartinR, I would say it doesn't because I can't use these powerful theorems yet. I thought Julio Puerta's was much better, but I also don't have this fundamental limit yet. – user1145880 Mar 29 '24 at 16:18
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    This answer https://math.stackexchange.com/a/2530893/42969 in that Q&A uses exactly your estimate. – Martin R Mar 29 '24 at 16:19
  • I agree. The second answer does. Thanks very much. (Should I delete the question? I wonder.) – user1145880 Mar 29 '24 at 16:21

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