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Check whether the series converge or diverge; $\sum_{n=1}^\infty(3^{\frac{1}{n^2}}-1)\csc^2(n).$

By the way, As $n\to \infty(3^{\frac{1}{n^2}}-1)\csc^2(n)=1.$ inconclusive.

I don't have an idea. How to check this series?

Any help will be appreciated. Thanks!

an4s
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  • Can you use the fact that $n^2(3^{1/n^2}-1)\to\log 3$? – Paramanand Singh Mar 25 '21 at 01:58
  • Is $\csc^2(n)=\frac{1}{\sin^2(n)}$? If yes, then determining the behavior of $\sum \frac{1}{n^2\sin^2(n)}$ isn't obvious to me. This post suggests that it's hard to do for an exponent of 3. Maybe it's easier for an exponent of 2. https://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1. – charlus Mar 25 '21 at 02:00

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The series is all positive terms, so no Alternating Series Test. The $\csc^2 (n)$ factor essentially oscillates chaotically between $1$ and $\infty$ so there is no point beyond which the terms stay reliably below a given $\epsilon$. The factor of $(3^{\frac{1}{n^2}}-1)$ goes to zero rapidly, but I don't think there is any limit to how high the cosecant factor can spike. So by the Divergence test, it diverges.

RobertTheTutor
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  • In the end it all depends at how frequently $n$ is close to $\pi/2\pm k\pi$, which depends itself on the irrationality of $\pi$. It's not obvious to me that which one dominates. – charlus Mar 25 '21 at 02:05
  • It doesn't have to dominate. The factor of $(3^{\frac{1}{n^2}}-1) can drop as quickly as it wants, but so long as the other factor is shooting to arbitrarily high values in an irrational pattern, the terms don't go to zero. Granted, I don't know how to prove that, but "trig function of integer" is pretty much a wild card. – RobertTheTutor Mar 25 '21 at 02:44
  • I think the problem is quite complicated (see my comment on the original post) – charlus Mar 25 '21 at 18:42