Concerning the sequence $\Big(\dfrac {\tan n}{n}\Big) $ Is the sequence $\Big(\dfrac {\tan n}{n}\Big) $ convergent ? If not convergent , is it properly divergent i.e. tends to 
either $+\infty$ or $-\infty$ ? ( Owing to $\tan (n+1)= \dfrac {\tan n + \tan 1}{1- \tan1 \tan n}$ and the non-covergence of 
$\Big (\tan n \Big)$ it is easy to see that if $\Big(\dfrac {\tan n}{n}\Big) $ is convergent then the limit must be  $0$ , but I can not proceed further . ) 

Since $\pi/2$ is irrational, a theorem of Scott says there exist infinitely many pairs of positive integers $(n,m)$ with $n$ and $m$ odd such that $\left| \dfrac{\pi}{2} - \dfrac{n}{m} \right| < \dfrac{1}{m^2}$. For such $m$ and $n$
we have $|\cos(n)| < |m \pi/2 - n| < 1/m$ and thus $|\tan(n)|/n > k$ for suitable nonzero constant $k$ (in fact any $k > 2/\pi$ should do).  So the sequence does not converge to $0$.   
