Is the sequence $n+\tan(n), n \in\mathbb{N}$ bounded below?

Intuitively I think it is not bounded below, but I have no idea how to prove it. It is like a Diophantine approximation problem, but most theorems seem to be too weak.

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    $\begingroup$ This might help math.stackexchange.com/questions/159438/… $\endgroup$ – leonbloy Jun 15 '18 at 3:12
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    $\begingroup$ Interesting question (+1). The numerators of the convergents of the continued fraction of $\frac{\pi}{2}$ are apparently good candidates to give negative results with large absolute magnitude. Not sure however, if this is already sufficient to prove the conjecture. $\endgroup$ – Peter Jun 16 '18 at 21:36
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    $\begingroup$ A spectacular example indicating that the sequence in fact seems to be unbounded : $$n=29973341381571568904696454415930130885472094353729$$ implies $$n+\tan(n)=-96430241781765207615806421837857889761609622768827.94009566\cdots $$ $\endgroup$ – Peter Jun 16 '18 at 21:45
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    $\begingroup$ $$3825751367438572721288064941140072998699766$$ $$982495809143557702771661080412376917282460965019427780239$$ gives a negative number with integer part having $101$ digits. $\endgroup$ – Peter Jun 16 '18 at 21:49
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    $\begingroup$ [Tangent Sequences, World Records, π, and the Meaning of Life: Some Applications of Number Theory to Calculus][1], page 372: If $\frac{p_k}{q_k}$ is a convergent for $\frac{\pi}{2}=[a1;a2,a3...]$ with $q_k$ odd then $\frac{|\tan p_k|}{p_k}$ is within $\frac23$ of $\frac{2(a_{k+1}+1)}{\pi}$. This roughly translates to $\frac{|\tan p_k|}{p_k}>2$ for $a_{k+1}\ge 4$ (which is pretty low) and $q_k$ odd (fairly common). The trick is that there is no easy way to tell if $\tan p_k$ is negative. [1]: jstor.org/stable/… $\endgroup$ – Oldboy Jun 19 '18 at 11:52

Each irrational number can be approximated through continuous fractions.

The result of each such approximation for the number $$\dfrac\pi2=[1; 1, 1, 3, 31, 1, 145, 1, 4, 2, 8, 1, 6, 1, 2, 3, 1, 4, 1, 5, 1, 41, 1, 2, 3, 7, 1, 1, 1, 27, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 49, 2, 1, 4, 3, 6, 2, 1, 3, 3, 17, 1, 3, 2, 1, ...]$$ can be represented in the form of $$\dfrac{P_{2i-1}}{Q_{2i-1}}<\dfrac\pi2<\dfrac{P_{2i}}{Q_{2i}},$$ where $$\dfrac{P_k}{Q_k}=\left\{\dfrac32,\dfrac{11}{7},\dfrac{344}{219},\dfrac{355}{226},\dfrac{51819}{32989},\dfrac{52174}{33215},\dfrac{260515}{165849},\dfrac{573204}{364913},\dfrac{4846147}{3085153},\dfrac{5419351}{3450066},\dfrac{37362253}{23785549},\dfrac{42781604}{27235615},\dfrac{122925461}{78256779},\dfrac{411557987}{262005952},\dfrac{534483448}{340262731},\dfrac{2549491779}{1623056876},\dfrac{3083975227}{1963319607},\dfrac{17969367914}{11439654911},\dfrac{21053343141}{13402974518},\dfrac{881156436695}{560961610149},\dfrac{902209779836}{574364584667},\dfrac{2685575996367}{1709690779483},\dfrac{8958937768937}{5703436923116},\dfrac{65398140378926}{41633749241295},\dfrac{74357078147863}{47337186164411},\dfrac{139755218526789}{88970935405706}\dots\right\}$$ wherein $$\left(\dfrac{P_{2i}}{Q_{2i}}-\dfrac{P_{2i-1}}{Q_{2i-1}}\right) = \dfrac1{Q_{2i-1}Q_{2i}},\quad 0< Q_{2i-1} < Q_{2i}.$$ Then $$\dfrac{P_{2i}}{Q_{2i}} - \dfrac1{Q_{2i-1}Q_{2i}}<\dfrac\pi2<\dfrac{P_{2i}}{Q_{2i}},$$ $$P_{2i} - \dfrac1{Q_{2i-1}}<\dfrac\pi2Q_{2i}<P_{2i},$$ $$\dfrac\pi2Q_{2i}<P_{2i}<\dfrac\pi2Q_{2i}+ \dfrac1{Q_{2i-1}},$$ $$P_{2i}=\dfrac\pi2Q_{2i}+ \dfrac\theta{Q_{2i-1}},$$ where $$\theta\in(0,1)\tag1.$$

If $Q_{2i}$ is odd, then $$\tan P_{2i}=\tan\left({\dfrac\pi2Q_{2i}+ \dfrac\theta{Q_{2i-1}}}\right) = -\cot{\dfrac\theta{Q_{2i-1}}} = -\dfrac1{\tan{\dfrac\theta{Q_{2i-1}}}}.$$ Taking in account the inequality $$\tan t \le \dfrac4\pi t,\quad t\in[0,1],\tag2$$ easy to get $$\tan P_{2i} < -\dfrac\pi{4\theta} Q_{2i-1}.\tag3$$ Therefore, if the conditions

  • $Q_{2i}$ is odd,
  • $\theta < \dfrac{2Q_{2i-1}}{P_{2i}},$

are satisfied for the infinite sequence of indexes $i,$ then for this sequence $$P_{2i} + \tan P_{2i} < -\left(\dfrac\pi2-1\right)P_{2i},$$ and, taking in account monotonic increase of the sequence $P_{2i},$ the issue sequence can not be bounded below.

On the other hand (without exact proof), the subsequence with the odd $Q_{2i}$ in the ratios \begin{align} &\dfrac{P_{2i}}{Q_{2i}}=\Bigg\{\mathbf{\dfrac{11}{7}},\dfrac{355}{226},\mathbf{\dfrac{52174}{33215}},\mathbf{\dfrac{573204}{364913}},\dfrac{5419351}{3450066},\\ &\mathbf{\dfrac{42781604}{27235615}},\dfrac{411557987}{262005952},\dfrac{2549491779}{1623056876},\mathbf{\dfrac{17969367914}{11439654911}},\mathbf{\dfrac{881156436695}{560961610149}},\\ &\mathbf{\dfrac{2685575996367}{1709690779483}},\mathbf{\dfrac{65398140378926}{41633749241295}},\dfrac{139755218526789}{88970935405706}\dots\Bigg\} \end{align} must be unlimited, and, taking in account $(1),$ the second condition, which uses the sequence \begin{align} &\dfrac{Q_{2i-1}}{P_2i}=\Bigg\{\mathbf{\dfrac{219}{11}},\dfrac{32989}{355}, \mathbf{\dfrac{165849}{52174}}, \mathbf{\dfrac{3085153}{573204}}, \dfrac{23785549}{5419351},\\ &\mathbf{\dfrac{78256779}{42781604}}, \dfrac{340262731}{411557987},\dfrac{1963319607}{2549491779}, \mathbf{\dfrac{13402974518}{17969367914}}, \mathbf{\dfrac{574364584667}{881156436695}},\\ &\mathbf{\dfrac{5703436923116}{2685575996367}}, \mathbf{\dfrac{47337186164411}{65398140378926}}, \dots\Bigg\} \end{align} really does not work.

So there are no reasons why the above conditions can make the required sequence limited.

This means that the sequence $n+\tan(n)$ is not bounded below.

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    $\begingroup$ Does this argument prove that for every irrational positive $\beta$, $n+\frac{1}{(2m+1)\beta-n}$ is unbounded below over $n,m\in\mathbb N$? $\endgroup$ – Henning Makholm Jun 23 '18 at 12:48
  • $\begingroup$ @HenningMakholm There is another task, which has independent answer. I have not this answer just now. My proof uses the specific features of the well-known number. $\endgroup$ – Yuri Negometyanov Jun 23 '18 at 15:47
  • $\begingroup$ @HenningMakholm Thanks, updated. $\endgroup$ – Yuri Negometyanov Jun 23 '18 at 19:16
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    $\begingroup$ I think everything is not clearly explained after you stated the inequality $\tan t\le\frac4\pi t$. Moreover, for the part without exact proof, how can you justify that the two conditions are satisfied infinitely many times simultaneously? If you cannot prove it, it would be better to state it as a conjecture. $\endgroup$ – Szeto Jun 24 '18 at 0:26
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    $\begingroup$ I agree with Szeto, the hard part is to show that the two conditions are satisfied infinitely often, this is highly nontrivial although heuristically probable. For this, you really have to know something general about convergents of $\pi/2$. $\endgroup$ – pisco Jun 25 '18 at 12:32

My answer is based on the conjecture(which is slightly different from the definition of irrationality measure) that, when $x=\frac{\pi}2$, for the inequality($p,q\in\mathbb{N}$) $$0<\frac{p}q-x<\frac1{q^{\mu(x)-\epsilon}}$$for every $\epsilon>0$, there exist infinitely many solutions $(p,q)$ with $q$ odd.

(Honestly, I believe this conjecture is true.)

Let’s firstly define a function $D(p)$ that measures the distance between $p$ and the nearest pole of $\tan$ on the left of $p$.

i.e. if:

  1. $x_0$ is a pole;
  2. $x_0<p$;
  3. $\tan$ is analytic in the interval $(x_0,p]$, then $D(p)=x_0$

For $p,q\in\mathbb{N}$, it can be shown that $$D(p)=p-(q\pi+\frac{\pi}2)$$ where $\frac{p}{\pi}-\frac32<q<\frac{p}{\pi}-\frac12$.

Rewritting a bit, $$D(p)=p-(q\pi+\frac{\pi}2)=(2q+1)\left(\frac{p}{2q+1}-\frac{\pi}2\right)\overbrace{=}^{Q=2q+1}Q\left(\frac{p}Q-\frac{\pi}2\right)$$

Conversely, if $\frac{x}{\pi}-\frac32<y<\frac{x}{\pi}-\frac12$ is true, then $D(x)=x-(y\pi+\frac{\pi}2)$.

This can be proved easily from observing the difference between the upper limit($\frac{x}{\pi}-\frac12$) and the lower limit($\frac{x}{\pi}-\frac32$) is $1$ and $y$ is an integer. Since difference between consecutive integers is $1$, every $x$ is unique to its $y$ and vice versa, and thus the converse is true.

Let $\mu$ be the irrationality measure of $\frac{\pi}2$. Let $\{(m,n)\}$ be the set of solutions to $(p,Q)$ for the inequality (call it $(1)$) $$0<\frac{p}Q-\frac{\pi}2<\frac1{Q^{\mu-\epsilon}}$$ (so $0<\frac{m}n-\frac{\pi}2<\frac1{n^{\mu-\epsilon}} $)

By the above conjecture, for every $\epsilon>0$, the set $\{(m,n)\}$ is infinite.

Trivially, it is also true that $$0<\frac{m}n-\frac{\pi}2<\frac{\pi}{n}$$if $\epsilon$ is sufficiently small.

With some simple algebra, this can be shown to be equivalent to $\frac{m}{\pi}-\frac32<n<\frac{m}{\pi}-\frac12$. Therefore, by the above 'converse theorem' $$D(m)=n\left(\frac{m}n-\frac{\pi}2\right)$$

Back to $(1)$, we get $$0<\frac{D}n<\frac1{n^{\mu-\epsilon}}\implies\color{BLUE}{0<D<\frac1{n^{\mu-\epsilon-1}}}$$

This implies $D$ can be arbitrarily small because $n$ can be arbitrarily big due to the infinite-ness of the set $\{m,n\}$. Plus, $D(p)$ measures the distance between $p$ and the left nearest pole; thus $p$ can be arbitrarily close to a pole from the right.

Next, the inequality $$\tan(m)<-\left(m-n\pi-\frac{\pi}2\right)^{-1+\delta}$$ is true for any $1>\delta>0$ and $D(m)$ sufficiently small. (Please recall that we have just proved $D$ can be arbitrarily small, in case you have forgotten.)


Also, we have shown that $$\frac{m}n-\frac{\pi}2<\frac{\pi}{n} $$ which implies $$m<\frac{\pi n}{2}+\pi$$

Together with the red inequality, we obtain $$\color{GREEN}{m+\tan(m)<-\left(\frac1{n^{\mu-\epsilon-1}}\right)^{-1+\delta}+\frac{\pi n}{2}+\pi}$$

Due to the set $\{(m,n)\}$ is infinite, $m,n$ can be arbitrarily big. If the first term on the right hand side is dominant, then $m+\tan( m)$ can be shown to be upper bounded by arbitrarily large negative numbers, which implies $m+\tan(m)$ is not lower bounded.

To get the dominance, we need $$(1-\delta)(\mu-\epsilon-1)>1\implies\mu>\frac1{1-\delta}+1+\epsilon$$meaning $\mu$ cannot be too close to $2$. Nevertheless, $\mu\left(\frac{\pi}2\right)$ is unknown. (I think this is quite likely because it makes sense that $\pi$ is slightly more irrational than $e$. There are many debates on the irrationality measure of $\pi$.)

In case $\mu\left(\frac{\pi}2\right)>2$, $m+\tan(m)$ is not lower bounded.

In case $\mu\left(\frac{\pi}2\right)=2$, whether $m+\tan(m)$ is lower bounded is not determined by the above method.


This is not a complete answer, just a summary of notes with a "visualisation" and insights into the problem (which I keep open for a couple of weeks and feel saddened to drop).

Note 1. From Kronecker's approximation theorem $$M=\left\{k\pi+n \mid k,n\in\mathbb{Z}\right\} \tag{1}$$ is dense in $\mathbb{R}$.

Note 2. Function $f(x)=x+\tan{(x)}$ is continuous on $\left(t\pi-\frac{\pi}{2},t\pi+\frac{\pi}{2}\right), \forall t\in\mathbb{Z}$ and has $\mathbb{R}$ as its range (easy to see by checking function's behaviour at $-\frac{\pi}{2}$ and $\frac{\pi}{2}$). Also $$f(-x)=-x+\tan{(-x)}=-x-\tan{(x)}=-f(x)$$

Note 3. It is clear that for $$\forall t_{k,n} \in M: f\left(t_{k,n}\right)=k\pi+n+\tan{(n)} \tag{2}$$ Now let's look at the cases when $f(x)\leq0$ enter image description here Of which, there are plenty for $x\leq0$ and getting pretty scarce for $x>0$.

Note 4. Because $M$ is dense in $\mathbb{R}$ (from Note 1) we can approximate any $x$ satisfying $f(x)\leq0$, e.g. $\left|x-t_{k_x,n_x}\right|<\delta$, and because $f(x)$ is continuous almost everywhere (from Note 2), $f\left(t_{k_x,n_x}\right)$ will approximate $f(x)$, e.g. $\left|f(x)-f\left(t_{k_x,n_x}\right)\right|<\varepsilon$. This means $$\forall x: f(x)\leq 0 \Rightarrow \exists t_{k_x,n_x}\in M: f(t_{k_x,n_x})\leq 0 \overset{(2)}{\Rightarrow} n_x+\tan{(n_x)}\leq-k_x \pi \tag{3}$$

Note 5. So far we established the existence of infinity of $n,k\in \mathbb{Z}$ s.t. $n+\tan{(n)}\leq-k \pi$. Obviously, if we want $n,k\in \mathbb{N}$, from Note 4, $x$ will have to be positive. But, from Note 3, these intervals are becoming pretty scarce, although not empty $$\left(t\pi-\frac{\pi}{2},\alpha_t\right]: \tan{(\alpha_t)}=-\alpha_t$$ and we want $k,n\in\mathbb{N}$ s.t. $$0<\color{red}{t\pi-\frac{\pi}{2}} < k\pi +n \leq \color{red}{\alpha_t}<t\pi \tag{4}$$ $t\in\mathbb{N}$, with the bounds $0\leq n<\pi\left(t-\frac{1}{2}\right)$ and $0\leq k < t-\frac{1}{2}$. It's not too dificult to show that

$$0<\alpha_t- t\pi+\frac{\pi}{2} \rightarrow 0, t \rightarrow \infty \tag{5}$$

From $$(4) \Rightarrow 0<\alpha_t- t\pi+\frac{\pi}{2}<\frac{\pi}{2} \Rightarrow -\frac{\pi}{2} < \alpha_t- t\pi < 0$$ which means $\lim\limits_{t\rightarrow\infty} \alpha_t \rightarrow \infty$ since $\lim\limits_{t\rightarrow\infty} t\pi \rightarrow \infty$. But $$\tan{\left(\alpha_t- t\pi\right)}=\tan{(\alpha_t)}=-\alpha_t \rightarrow -\infty, t \rightarrow \infty$$ which is only possible when $\alpha_t- t\pi \rightarrow -\frac{\pi}{2} \Rightarrow 0<\alpha_t- t\pi+\frac{\pi}{2} \rightarrow 0, t \rightarrow \infty$.

Note 6. In fact (forced by $(5)$), we want $(4)$ as small as possible

$$0<\color{red}{t\pi-\frac{\pi}{2}} < k\pi +n \leq \color{red}{t\pi-\frac{\pi}{2}+\varepsilon} \iff \\ 0<2t\pi-\pi < 2k\pi+2n \leq 2t\pi-\pi+2\varepsilon \\ 0<2t\pi-\pi - 2k\pi < 2n \leq 2t\pi-\pi-2k\pi+2\varepsilon$$ $$\frac{\pi}{2} < \frac{n}{2(t-k)-1} \leq \frac{\pi}{2}+\frac{\varepsilon}{2(t-k)-1} \tag{6}$$ leading to odd denominators and best rational approximations of $\frac{\pi}{2}$ with odd denominators (and even A046965 and Newton/Euler) explored by the other answers.


I think that the ergodicity of $\tan x$ could be used to prove that $n+\tan n$ is unbounded. I am not able to prove it exactly, so I leave this answer as a community wiki.

  • $\begingroup$ I left this answer above: "I think that the ergodicity (ams.org/journals/proc/1978-071-01/S0002-9939-1978-0473144-4/…) of $\tan x$ coul be used to prove that $n+\tan n$ is unbounded. But I am not able to prove it exactly, that's why this answer is a community wiki. $\endgroup$ – zar Jun 21 '18 at 22:30
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    $\begingroup$ This could have been a comment. $\endgroup$ – Sungjin Kim Jun 21 '18 at 23:55
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    $\begingroup$ I opened the answer as a community Wiki in order to let experts to expand it from a comment to a complete answer $\endgroup$ – zar Jun 22 '18 at 5:25
  • $\begingroup$ I have the feeling that probably the most that the ergodicity theorem could prove is something along the lines of: for almost every $x$, $\tan (nx) + nx$ is unbounded. But that wouldn't help in trying to prove the statement for any particular $x$. $\endgroup$ – Daniel Schepler Jun 22 '18 at 16:38
  • $\begingroup$ @DanielSchepler $n+ \tan n$ is unbounded: For infinitely many $n,$ both $\cos n,\sin n$ are positive. Along this subsequence of $n$'s we have the expression $\to \infty.$ $\endgroup$ – zhw. Jun 22 '18 at 19:55

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