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I'm wondering whether the following series

$$\sum_{k=1}^{\infty}\frac{\cos\sqrt{k}}{k}$$

is convergent or not. Any hint will be appreciated.

SimonChan
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  • conditionally convergent b/c alternating series and $$\lim_{k \to \infty} = 0$$ – Christopher Marley Jan 31 '19 at 04:48
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    Why is it alternating? @ChristopherMarley – Clement C. Jan 31 '19 at 04:49
  • @ClementC. the $\cos$ term is alternating between positive and negative, albeit, seemingly random values. The $\sqrt(k)$ term ensures that the series will alternate because it's never a multiple of pi (for integer $k$s). Therefore, $\cos\sqrt k$ is alternating – Christopher Marley Jan 31 '19 at 04:52
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    That's not what alternating means. – anomaly Jan 31 '19 at 04:53
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    To apply Leibniz's criterion, you need consecutive terms to be of different sign. This is not the case here. What you call "alternating" is not what alternating is supposed to mean. @ChristopherMarley – Clement C. Jan 31 '19 at 04:54
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    @anomaly My bad, I clearly wasn't thinking. I forgot about "alternating." – Christopher Marley Jan 31 '19 at 05:01
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    The derivative of square root goes to 0 so you could possibly approximate with an integral – Mark Jan 31 '19 at 05:22
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    https://math.stackexchange.com/questions/2948394/does-sum-n-1-infty-frac-cos-sqrtnn-converge – G. Smith Jan 31 '19 at 05:28
  • @Mark hey man. What do you mean? Can you explain your thoughts a little bit more or maybe provide a link to related theorem? – Levon Minasian Feb 22 '21 at 18:29
  • @LevonMinasian What I mean is that, for big enough $n$, $\int_n^{n+1} \cos(\sqrt x) /x dx$ is very close to $\int_n^{n+1} \cos(\sqrt n) / x dx$ so it could be massaged into a bound for the sum – Mark Feb 23 '21 at 01:17
  • @LevonMinasian I elaborated here https://math.stackexchange.com/a/4038108/4460 – Mark Feb 24 '21 at 13:21

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