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The one in question is as followed:

$$ \sum_{n=1}^{\infty} e^{-n}\tan n $$

This comes up in one of those coffee-shop discussion...

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  • $\begingroup$ I don't know how to make it rigorous, but I'm inclined to guess it diverges from a probabilistic argument: each congruence class of subintervals of length $\pi/k$ of $(-\pi/2, \pi/2)$ occurs equally often, and so the magnitude of contribution from each is $c \int_a^{a+\pi/k} \tan(t) dt$ for some constant $c$, which goes to $\infty$ as $a \to 0$. $\endgroup$ – David Schneider-Joseph Aug 2 '17 at 3:49
  • $\begingroup$ My guess is that it would converge since $exp(-n)$ gets nicely small while $tan(n)$ depends on how close $n$ is to $(k+1/2)\pi$ for some $k$, and this would not happen too often. $\endgroup$ – marty cohen Aug 2 '17 at 4:10
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    $\begingroup$ Using the fact that $\pi$ has finite irrationality measure, you can check that $\tan n$ grows at most polynomially fast. So the series converges absolutely. $\endgroup$ – Sangchul Lee Aug 2 '17 at 5:07
  • $\begingroup$ @SangchulLee: Can you please add an answer elaborating it? I am unable to see how. $\endgroup$ – Aryabhata Aug 2 '17 at 5:44
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For each $x \in \mathbb{R}$, its irrationality measure $\mu(x)$ is defined as

$$ \mu(x) = \inf \left\{ c \in \mathbb{R} : \left| x - \frac{a}{b} \right| \leq \frac{1}{|b|^c} \text{ for at most finitely many } (a, b) \in \mathbb{Z}\times\mathbb{Z}^{*}\right\}. $$

Let us collect some basic properties of $\mu$:

  1. It is clear that $\mu(x) = 1$ if $x$ is rational. On the other hand, if $x$ is irrational, then by Dirichlet's approximation theorem we have $\mu(x) \geq 2$.

  2. We have $\mu(x) = \mu(1/x)$. Indeed, it suffices to prove this for irrational $x$. To this end, notice that if $0 < c < \mu(x)$, then there exists $(a_j, b_j) \in \mathbb{Z}^* \times \mathbb{Z}^*$ such that $|b_j| \to \infty$ and that $|x - (a_j/b_j)| \leq |b_j|^{-c}$. In particular, $a_j/b_j \to x$. Then it follows that

    $$ \left| \frac{1}{x} - \frac{b_j}{a_j}\right| = \frac{|b_j/a_j|}{|x|} \left| x - \frac{a_j}{b_j} \right| \leq \frac{\text{const}}{|b_j|^c} \leq \frac{\text{const}}{|a_j|^c} \leq \frac{1}{|a_j|^{c-\epsilon}} $$

    if $\epsilon > 0$ and $j$ is sufficiently large. So we have $c-\epsilon \leq \mu(1/x)$ and this is enough to conclude the claim.

  3. It is well-known that $\mu(\pi) < \infty$.

Using these properties, we find that $\mu(1/\pi) < \infty$. So we can pick $c > \mu(1/\pi)$. Then

$$ |\cos n| = \left|\sin\pi\left(\frac{n}{\pi} - \frac{1}{2} - a \right) \right| $$

for any $a \in \mathbb{Z}$. Pick $a$ such that $\left| \frac{n}{\pi} - \frac{1}{2} - a \right| \leq \frac{1}{2}$. Then using the inequality $|\sin(\pi x)| \geq 2|x|$ for $|x| \leq \frac{1}{2}$, for large $n$ we have

$$ |\cos n| \geq |2n|\left|\frac{1}{\pi} - \frac{2a+1}{2n} \right| \geq |2n|\cdot\frac{1}{|2n|^{c}} = \frac{1}{|2n|^{c-1}}. $$

This shows that $|\tan n| \leq |2n|^{c-1}$ for large $n$ and hence the series converges absolutely by comparison test.

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This does not prove anything.

Sharing the same opinion as Marty Cohen, I just performed numerical evaluations of $$S_k=\sum_{n=1}^{10^k}e^{-n}\, \tan(n)$$ and obtained the following results $$\left( \begin{array}{cc} k & S_k \\ 1 & 0.2663227174666176381276531 \\ 2 & 0.2625520215463125833311472 \\ 3 & 0.2625520215463125833311472 \\ 4 & 0.2625520215463125833311472 \\ 5 & 0.2625520215463125833311472 \\ 6 & 0.2625520215463125833311472 \end{array} \right)$$

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  • $\begingroup$ Using the known bounds for the irrationality measure of $\pi,$ it's possible to prove $|\tan n|\ge n^C$ for sufficiently large $n$ and some $0<C<8$, that's sufficient for convergence. Ah, I see that @Sangchul Lee wrote that, already. $\endgroup$ – Professor Vector Aug 2 '17 at 6:08

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