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I have arrived at the point where I have a series

$$\sum_{n=1}^{\infty}\frac{\sin^2(n)}{2n}$$

that I know should diverge (checking via python implies it might be divergent, and wolframalpha times out when trying to evaluate it, although yes, it doesn't truly mean anything).

I have tried multiple approaches and convergence tests, but most of them appear to not work with alternating series. Comparing to a harmonic series also didn't lead me anywhere.

Literally out of ideas, does anyone have an approach to this I might be missing?

artoftheblue
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  • @ericw31415 I need to find some sequence that would be less than sin^2(n)/(2n), and I can't estimate it the same way as I could with convergent series (since there I would try to find a sequence that would be more than sin^2(n)/(2n), so 1/2n, for instance, since sin^2(n) has an upper bound of 1) – artoftheblue Mar 08 '24 at 20:54
  • @IzaakvanDongen it appears to, does the fancy R in that question mean the real part of the complex number? I think I would be able to take it from there if it so -- much appreciated! – artoftheblue Mar 08 '24 at 20:58
  • Yes, that's right! Really the key idea is that the trig identity lets you decompose the sum into two parts, a harmonic part, and another part which converges. Sums similar to the other part are discussed here, here, here and here as well, if you would like some more details/alternative techniques/more general results. – Izaak van Dongen Mar 08 '24 at 21:18
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    Intuition: $\sin^2(n)$ is always positive, and for $n$ sufficiently close (say within $\delta$ for some $\delta > 0$) to values $\pm \frac{\pi}{2} + 2\pi k$ it's bounded below (say by $\alpha_\delta > 0$). Then integers $n$ will be within $\delta$ of $\pm \frac{\pi}{2} + 2 \pi k$ at least once in a period $A$ bounded above only by the size of $\delta$. Thus this sum is at least $\sum \frac{\alpha_\delta}{An} = \frac{\alpha_\delta}{A} \sum \frac{1}{n}$, and hence diverges. I'm implicitly using a pigeonhole argument for remainders of $n$ divided by $\pi$. Stronger equidistribution is known. – davidlowryduda Mar 08 '24 at 21:31

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