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I tried searching mathematical literature but I was unable to come across anything apart from the non-convergence of the sequence.

Simply put:

Define $\displaystyle A = \left.\left\{\frac{\tan\left(n\right)}{n}\,\right\vert\, n \in \mathbb{N}\right\}.\quad$ Is $\overline{A} = \mathbb{R}$ ?.

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  • $\begingroup$ What makes you think that $\overline A = \mathbb R$? $\endgroup$
    – lisyarus
    Jun 19 '19 at 16:43
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    $\begingroup$ Just a hunch. Nothing rigorous. $\endgroup$ Jun 19 '19 at 16:44
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    $\begingroup$ I am not asking for a rigorous proof; I am asking about what drives your hunch, - it may help others answer your question. $\endgroup$
    – lisyarus
    Jun 19 '19 at 16:51
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    $\begingroup$ I suspect that the convergents of π/2 approach π/2 "faster" than n grows, for some well defined notion of "faster". Hence the sequence jumps to arbitrary points in ℝ, and cover it densely. But this is all founded on nothing more than a moment's thought. $\endgroup$ Jun 19 '19 at 16:54
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tl;dr (written after addendum): It seems plausible that $A$ contains points of both signs having arbitrarily large magnitudes. It's not so clear $A$ is dense outside of a neighborhood of zero.


Suppose that $\varepsilon = \left| \dfrac{\pi}{2} - \dfrac{n}{k} \right|$ is small, so that $n$ is nearly $k \pi/2$. If $k$ is even, $\dfrac{\tan n}{n}$ is small. If $k$ is odd, $\dfrac{\tan n}{n}$ is large. Dirichlet's approximation theorem tells us there are infinitely many $n,k$ pairs with $\varepsilon < \dfrac{1}{k^2}$.

By applying Dirichlet's approximation theorem to $\pi$, and arguing that this new set of pairs is a subset of the above set of pairs, we find there are infinitely many even $k$. But we do not know if there are infinitely many odd $k$.

Using the first $50\,000$ convergents of the continued fraction for $\pi/2$ as $\dfrac{n}{k}$ pairs, and plotting $\left( \log_{10} n,\dfrac{\tan n}{n} \right)$, we have

Plot emphasizing outliers

and zooming in the vertical axis a bit

Plot showing detail closer to the x-axis.

there do seem to be large magnitude outliers and no tendency to collapse towards zero, at least up to $n$ near $10^{25\,000}$. (Note that values of $\left| \dfrac{\tan n}{n} \right| <10$ are suppressed to avoid a solid band of points obscuring the horizontal axis.)


Extending a bit further, to the first $200\,000$ convergents... These only plot points for odd $k$, which is $133\,363$ of the convergents (so maybe there is a significant bias toward odd $k$s).

extreme values out to a little past 10^100000

smaller near- odd multiple of pi values out to a little past 10^100000

Maybe there's a small bias favoring positive extreme values. We don't seem to run out of new extreme values. There was no need to cut out small magnitude points since we cut out even $k$s. We don't seem to be running out of odd $k$s, so maybe there are infinitely many of them. The vertical density of points seems stationary as $n$ increases. But there also seems to be very little reason to believe in eventual fill-in of the gaps between the very large magnitude points.

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