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)

  • $\begingroup$ really it is zero? $\endgroup$
    – user
    Dec 5, 2017 at 18:51
  • $\begingroup$ @gimusi The continuous limit has infinite lim sup and -infinite lim inf, and I doubt that the discrete limit exists $\endgroup$
    – Coolwater
    Dec 5, 2017 at 18:51
  • $\begingroup$ math.stackexchange.com/questions/686841/… $\endgroup$
    – Guy Fsone
    Dec 5, 2017 at 18:52
  • 2
    $\begingroup$ I imagine this is going to depend on the irrationality measure of $\pi$. I'm sure there is such a $k$ where the limit goes to 0 because $\pi$ is not a Louisville number, but I don't know if it's 8. Where do you get this constant? $\endgroup$
    – user123641
    Dec 5, 2017 at 18:54
  • 1
    $\begingroup$ @p.kn I was talking as if $n$ were a natural number. $\endgroup$
    – user123641
    Dec 5, 2017 at 19:21

1 Answer 1


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 $k=3$.

  • $\begingroup$ that's incredible! $\endgroup$
    – user
    Dec 5, 2017 at 21:39
  • $\begingroup$ Can you provide the proof of "while for $k>2$ is goes rapidly to zero." $\endgroup$ Dec 5, 2017 at 23:42
  • $\begingroup$ @Mark Fischler, thanks. as i noticed you said that if $\frac{p}{q}$ is continued fraction approximation of $\frac{\pi }{2}$ then by taking $n$ as denominator of $\frac{p}{q}$, $\tan (n) \approx \frac{1}{q}$. could you please give the proof of this statement.but, why do you used the reciprocal of difference of two continued fraction as the approximation of tan(n) in your code instead of what you said. $\endgroup$
    – p.kn
    Dec 6, 2017 at 6:53

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