Does the sum $\sum_{n=1}^{\infty} e^{-n}\tan n$ converge at all? 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...
 A: 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 $\mu$:


*

*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 $\mu(x) \geq 2$.

*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.

*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.
A: 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
$$\left(
\begin{array}{cc}
 k & S_k \\
  1 & 0.2663227174666176381276531 \\
 2 & 0.2625520215463125833311472 \\
 3 & 0.2625520215463125833311472 \\
 4 & 0.2625520215463125833311472 \\
 5 & 0.2625520215463125833311472 \\
 6 & 0.2625520215463125833311472
\end{array}
\right)$$
