Sequence $\tan(n)/n$ Yes I know the function $f(x) = \frac{\tan(x)}{x},\; x \in\mathbb{R}$ does not converge for $x\rightarrow \infty$.
But what about the sequence where you only take integers for $n$. If I would guess, I would say it does not converge because some Integers might be 'near enough' to one of the pols to 'overcome' the $\frac{1}{n}$. 
But either way I can't proof it. Has anybody an Idea? 
Thank you for your answers in advance.
 A: Since $\pi$ is an irrational number, by the Lagrange's theorem there is an infinite number of rational numbers $\frac{p_n}{q_n}$ such that
$$ \left |\pi-\frac{p_n}{q_n}\right|\leq \frac{1}{q_n^2},$$
and the previous line implies:
$$ \left| \sin(p_n)\right| \leq \frac{1}{q_n} $$
so there is an infinite number of natural numbers $m$ such that the absolute value of $\frac{\tan m}{m}$ is very close to $\frac{1}{\pi}$ (or even bigger). A subsequence of $\left\{\frac{\tan m}{m}\right\}_{m\geq 1}$ is clearly convergent to zero, hence it follows that such a sequence is not a Cauchy sequence. Moreover, since the general term of every convergent series has to be infinitesimal, we have that
$$ \sum_{m\geq 1}\frac{\tan m}{m} $$
is not convergent. On the other hand, the series
$$ \sum_{m\geq 1}\frac{\tan m}{m^\alpha} $$
is convergent for any $\alpha>\alpha_0$, but the value of $\alpha_0$ depends on the irrationality measure of $\pi$, whose exact value is still unknown - it is conjectured to be $2$, but no one ever proved something better than $7.6$.
