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.
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.
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
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.
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:
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.)
$$\color{RED}{\tan(m)<-\left(m-n\pi-\frac{\pi}2\right)^{-1+\delta}=-D^{-1+\delta}<-\left(\frac1{n^{\mu-\epsilon-1}}\right)^{-1+\delta}}$$
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 $f(x)\leq0$ 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.