Devise strategy to hit submarine (Based on a problem from Brilliant).
Suppose there is an enemy submarine at unknown location on real number line moving at unknown real-valued  velocity. You can fire one missile per minute in attempt to hit the submarine, and your missile will hit any submarine within fixed radius $\varepsilon>0$. Is there a strategy to guarantee you hit the submarine in a finite number of steps?
Note that if the submarine has integer position and integer velocity, then we can simply enumerate all position-velocity pairs $(x_1,v_1),(x_2,v_2),(x_3,v_3),\dots$ (since $\Bbb Z^2$ is countable) then on the $n$-th step, fire your missile at position $x_n+nv_n$, which is where the $n$-th submarine would be at step $n$. For integer-valued position and velocity, this is a winning strategy.
For real-valued position and velocity, the problem can be formalized like this: Let $\varepsilon>0$. Does there exist a function $f:\Bbb N\to\Bbb R$ such that for every linear function $s:\Bbb N\to\Bbb R$, given by $s(n)=x+nv$ for $x,v\in\Bbb R$, there exists $n\in\Bbb N$ such that $|f(n)-s(n)|<\varepsilon$?

Attempt. For each $\varepsilon>0$, $n\in\Bbb N$, $f\in\Bbb R$, the set $$\{(x,v)\in\Bbb R^2:|f-(x+nv)|<\varepsilon\}$$
(the set of all submarines you kill by firing at position $f$ and step $n$)
forms a thin band in $\Bbb R^2$. The question is whether for $n=1,2,3,\dots$ we can generate sequence of bands that cover the entire $\Bbb R^2$ (thus hitting every possible submarine, a winning strategy).
Observe that the width of the bands $\to0$ as $n\to\infty$, but the width is similar to $1/n$ so it's like the harmonic series which is not finite. This provides good hope that these bands can cover $\Bbb R^2$ but I cannot dream of a sequence $f$ which guarantees this.
 A: It seems to me it is possible to cover $\mathbb{R}^2$ using a sequence of strips $(S_n)_n$ of the form
$$S_n:=\{(x,v)\in \mathbb{R}^2| \,|f-(x+nv)|<\varepsilon\}$$
wherein $f$ is the only free parameter at our disposal. Let me sketch a would-be proof. Of course, I can certainly cover the ball $B(0,\varepsilon)\subset \mathbb{R}^2$ and suppose that I have even managed to cover $B(0,m\varepsilon)$ with the first $M(<+\infty)$ members of the sequence $(S_n)_n$. Can I then cover the surrounding annulus $B(0,(m+1)\varepsilon)\setminus B(0,m\varepsilon)$ with a finite number of additional members in the remaining tail $(S_{M+1},S_{M+2},S_{M+3},...)$? That seems possible. Speaking graphically, I color this annulus from top to bottom, making sure that I do not omit any uncolored region along the way: first I have the upper cap of the annulus rigorously colored, then the left part, then the lower cap and finally I fill up any ommisions in the right part of the annulus. Will this process finish after a finite number of "paint-strokes"? The orientation of the paint-strokes is given and approximately horizontal while their width decays as $1/n$. So with every stroke I can approximately proceed downward by a distance $1/n$ which, by virtue of the divergence of the harmonic series, suffices to finish the job. So, for a certain $M'\in \{M+1,M+2,...\}$ I manage to cover the annulus $B(0,(m+1)\varepsilon)\setminus B(0,m\varepsilon)$ with the strips $(S_{M+1},...,S_{M'})$. In other words, I have covered $B(0,(m+1)\varepsilon)$ using the first $M'$ members of the sequence $(S_n)_n$. Continuing this algorithm inductively on $m$ eventually yields a complete covering of $\mathbb{R}^2$.
N.B. In 2 spatial dimensions you can no longer hit the submarine with certainty. In that case, the slabs
$$S_n = \{(x,y,v_x,v_y)\in \mathbb{R}^4| (f_x-(x-nv_x))^2+(f_y-(y-nv_y))^2<\varepsilon^2\}$$
have a too small Lebesgue-volume in the following sense: there is a constant $C>0$ s.t. the volume of $S_n \cap B(z,R)$ is bounded by $C\frac{R^2}{n^2}$. Suppose that I've already used the first $N$ slabs in my attempt to cover $\mathbb{R}^4$ and $N$ is so large that $\sum_{n=N+1}^\infty \frac{C}{n^2}< \frac{\pi^2}{2}=Vol(B(0,1))$. During this intermediate stage of the attempt to cover $\mathbb{R}^4$, I can still find a yet completely uncovered ball $B(z_0,1)$ somewhere. However,
$$Vol(B(z_0,1))=\frac{\pi^2}{2}>\sum_{n=N+1}^\infty Vol(S_n\cap B(z_0,1)),$$
so that I'll be unable to completely cover that ball.
A: W.l.o.g. let $\epsilon=2$ and assume the initial position and initial velocity are positive. 
At minute $n$, take a coin of height $\frac{1}{n}$cm. Use the coins to create infinitely many stacks, at positions 0,1,2..., following the strategy: first make the first stack at least height 1cm, then make the first two stacks at least height 2cm, and so on. 
It's easy to show that you can continue indefinitely with this (*), hence each stack will become arbitrarily high. 
This gives the solution to your problem: if coin $n$ goes to stack $m$ and that stack is size $v$ before you add the new coin, fire a missile at position $m+nv$. The space taken by the coin in the two dimensional stacks of coins plane is the part of the initial configurations that you ruled out in that step.
(*) Suppose you conclude step $k$ of the strategy at time $n_k$.  To conclude step $k+1$, you need another $(k+k+1)$cm worth of coins, and since  $\sum_{n \geq n_k} 1/n =\infty$ you will eventually get them. 
