Does the sequence $n+\tan(n), n \in\mathbb{N}$ have a lower bound? 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.
 A: 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: 


*

*$x_0$ is a pole; 

*$x_0<p$; 

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

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.
A: 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.
A: 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.
