$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\tan (n)}}{{{n^k}}}$ what is the minimum number of $k$ for which the following limit exist
$$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\tan (n)}}{{{n^k}}}$$ 
I know that  $$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\tan (n)}}{n}$$
doesn't exist, and $$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\tan (n)}}{{{n^8}}} = 0.$$
But i don't know what is the minimum  number of $k$ for existing that limit.
(note that here n's are positive integers not real numbers)
 A: The practical way to approach this is as follows:  Consider the $m$-th continued fraction approximant for $\pi/2$.  (There is a conjecture that the denominator $q_m$ approaches  $L^m$ where $L$ is Levy's constant (which is about $3.28$), but I can't see how to exploit that conjecture here.)
These approximants are the localized "best" you can do relative to the size of the denominators, so if $\frac{\tan n}{n^k}$ diverges then it will also diverge considering only those values of $n$ which are denominators of those continued fractions.
The error in each approximation is roughly the difference between that approximation and the next one; in fact, because the signs of the errors are alternately positive and negative, this error estimate provides an upper bound for the absolute value of the error.  But then when odd $n$ is a continued fraction denominator, $\tan n$ is (in absolute value) very nearly the reciprocal of the continued fraction denominator.
Then it is straightforward to examine the "trends" for various exponents $k$ by dividing that approximation to $\tan n$ by $n^k$ for just those values of $n=q_m$ as $m$ grows.  
When you do this, you see that the boundary case (treating $k$ as a real number) is at $k=2$: for $k<2$ this sequence of excellent approximations to $\frac{\tan n}{n^k}$ quickly grows very large, while for  $k>2$ is goes rapidly to zero.
When $k=2$ the values of $\frac{\tan n}{n^k}$ are distributed between about $\pm 16$, with slightly more density near $\pm 1$ but a gap between $-1$ and $+1$ which can be explained based on the nature of the approximations taken.  At any rate, it is "clearly" not going to zero.
If you have Mathematica you can plot these:

fracs := Table[
    FromContinuedFraction[ContinuedFraction[Pi/2, m]], {m, 1, 1000}]
fracsR := Take[fracs, {2, 1000}]
diffs := Take[fracs, {1, 999}] - fracsR
ListPlot[N[(1/diffs)/Denominator[Take[fracs, {1, 999}]]^2]]

So the answer is that the first integer $k$ for which $\lim_{n\to\infty}\frac{\tan n}{n^k} = 0$ is $k=3$.
