I was wondering weather the series $$\sum_{n=1}^{\infty} \frac {\tan(n)} {n^b}$$ diverges or converges whenever $b \geq 1$ is an integer. Does anyone have a proof for either statement? Does it converge for some positive integers but not for others? In that case, which are they?

  • 2
    $\begingroup$ It is closely related to the irrationality measure $\mu$ of $\pi$, as $\mu$ determines the speed at which $\tan n$ grows. Indeed, it is well-known that $\mu < \infty$, so that the series converges for large $b$. $\endgroup$ – Sangchul Lee Mar 1 '14 at 23:09
  • $\begingroup$ See math.stackexchange.com/questions/470527/… $\endgroup$ – dani_s Mar 1 '14 at 23:10
  • $\begingroup$ @dani_s Thanks, this one was linked: jstor.org/discover/10.2307/… I dont feel like signing up in order to read the paper, do you know if the paper is relevant? In that case, can it be found somewhere else? $\endgroup$ – user117449 Mar 1 '14 at 23:16
  • $\begingroup$ @sos440, do you have any evidence for the series to converge for large $b$? How large? $\endgroup$ – user117449 Mar 1 '14 at 23:20
  • 1
    $\begingroup$ Any $b > \mu(\pi^{-1}$$ works. $\endgroup$ – Sangchul Lee Mar 1 '14 at 23:31

The series certainly diverges for $b=1$ and converges for $b\ge 8$, as shown in this article by Sam Coskey. Moreover, it "stays very small for a very long time" when $b=2$.

In one of the answers to this question, it is argued that the series will converge if $b > \mu(\pi^{-1})$, where $\mu$ is the irrationality measure. The irrationality measure of $1/\pi$ is the same as the irrationality measure of $\pi$, which has recently received an improved upper bound of $7.6063$ (Salikhov 2008; see MathWorld's article for full reference). While this argument reproduces the already-known result (convergence for $b\ge 8$), it shows that a tighter upper bound on $\mu(\pi)$ would immediately imply convergence for smaller values of $b$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy