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 . ) 

If you plot $(\tan x)/x$, e.g. by typing "(tan x)/x" into Google or Wolfram Alpha, you'll see a plot like this (the below is from Google):

So you can see clearly that $\frac{\tan x}{x}$ has no limit as $x \to \infty$, nor does it diverge to $\pm \infty$, as it oscillates. Indeed it infinitely often takes the value $0$ (for $x = k\pi$ for integers $k$) and infinitely often approaches $\pm \infty$ (for $x$ around $k\pi \pm \pi/2$ for integers $k$).

For the sequence $(\tan n)/n$ where $n$ only ranges over integer values, we should expect the same sort of behaviour, considering that the period $\pi$ of $\tan x$ is irrational and so there is nothing special about restricting the sequence to integers. [On the other hand, taking only multiples of $\pi$ would be special, e.g. $(\tan (\pi x))/x$ has a similar plot – doesn't converge – but clearly $(\tan (\pi n))/n = 0$ for all integers $n$.]
For example, for positive $n$, we can see that 


*

*$(\tan n)/n > 0$ (because $\tan n > 0$) when $2k\pi < n < (2k+1)\pi$ for some integer $k$ (i.e., when the integer $\lfloor n/\pi \rfloor$ is even), and 

*$(\tan n)/n < 0$ (because $\tan n < 0$) when $(2k+1)\pi < n < (2k+2)\pi$ for some integer $k$ (i.e., when the integer $\lfloor n/\pi \rfloor$ is odd).


As both of them happen infinitely often ($\lfloor n/\pi \rfloor$ takes on all nonnegative integer values as $n$ varies over the positive integers, as $\pi > 1$, so $\lfloor (n+1)/\pi \rfloor - \lfloor n/\pi \rfloor \le 1$, i.e., no integer can be "skipped"), this means that $(\tan n)/n$ takes on positive and negative values infinitely often.
This would be almost enough, except that a sequence taking on both positive and negative values can still have limit $0$, which we need to show is not possible.
For this, we can show that it infinitely often takes values bounded away from $0$. This would involve proving that infinitely often, $n$ is sufficiently close (how close is "sufficient" is a function of $n$) to $k\pi \pm \pi/2$ for some integer $k$, or equivalently that the fractional part of $n/\pi$ is sufficiently close to $\frac12$. Ultimately, this hinges on the irrationality of $\pi$: for instance, it is a theorem (equidistribution theorem) that the fractional parts of the sequence $n/\pi$ are uniformly distributed in $(0, 1)$.
More precisely, to show that infinitely often $|\tan n|/n > c$ happens, we need to find infinitely many $n$ such that $|\tan n| > cn$. We can get some crude lower bounds on $|\tan n|$ for $n$ near $k\pi \pm \pi/2$. We have, for $0 < \epsilon < \pi/2$, the fact that $|\cos \epsilon| \ge 1 - \epsilon^2/2$ and $|\sin \epsilon| \le \epsilon$, so 
$$|\tan (\pi/2 \pm \epsilon)| = \frac{|\cos \epsilon|}{|\sin \epsilon|} \ge \frac1\epsilon - \frac\epsilon2 > \frac1\epsilon - 1.$$ 
So it is enough to find infinitely many $n$ for which, with $\epsilon = \pi|\{n/\pi\} - 1/2|$ (where $\{n/\pi\}$ denotes the fractional part of $n/\pi$), we have $\frac1\epsilon - 1 > cn$, or equivalently that $\epsilon < 1/(1 + cn)$. In other words, it is enough to find infinitely many $n$ such that the fractional part of $n/\pi$ lies in $$\left(\frac12 - \frac{1}{\pi(1+cn)}, \frac12 + \frac{1}{\pi(1+cn)}\right),$$ for any $c > 0$ (e.g. $c = 1$).
Robert Israel's answer outlines an approach to this hardest part of the problem.
So $(\tan n)/n$ does not converges, nor does it diverge to $\infty$ nor to $-\infty$.
