Putting many disks in the unit square 
Consider a square of side equal to $1$. Prove that we can place inside the square a finite number of disjoint discs, with different radii of the form $1/k$ with $k$ a positive integer, such that the area of the remaining region is at most $0.0001$.

If we consider all the discs of this form, their total area is $\sum_{k \geq 1}\displaystyle \pi \frac{1}{k^2} - \pi=\frac{\pi^3}{6}-\pi\simeq 2.02$ which is greater than the area of the square. (I subtracted $\pi$ because we cannot place a disc of radius $1$ inside the square).
So the discs of this form can cover the square very well, but how can I prove that there is a disjoint family which leaves out a small portion of the area?
 A: Let's roll up our sleeves here. Let $C_k$ denote the disk of radius $1/k$. Suppose we can cover an area of $\ge 0.9999$ using a set of non-overlapping disks inside the unit square, and let $S$ denote the set of integers $k$ such that $C_k$ is used in this cover.
Then we require
$$\sum_{k\in S}\frac{1}{k^2} \ge 0.9999/\pi \approx 0.318278$$
As the OP noted, we know that $1 \not\in S$. This leaves
$$\sum_{k\ge2}\frac{1}{k^2} \approx 0.644934$$
which gives us $0.644934 - 0.318278 = 0.326656$ 'spare capacity' to play with.
Case 1 Suppose $2 \in S$. Then the largest disk that will fit into the spaces in the corners left by $C_2$ is $C_{12}$, so we must throw $3,...,11$ out of $S$. This wastes
$$\sum_{k=3}^{11}\frac{1}{k^2}\approx0.308032$$
and we are close to using up our spare capacity: we would be left with $0.326656-0.308032=0.018624$ to play with.  
Case 2 Now suppose $2 \not\in S$. Then we can fit $C_3$ and $C_4$ into the unit square, but not $C_5$. So we waste
$$\frac{1}{2^2} + \frac{1}{5^2} = 0.29$$
leaving us with $0.326656-0.29=0.036656$ to play with.  
Neither of these cases fills me with confidence that this thing is doable.
A: I suggest the following idea:
Look at http://en.wikipedia.org/wiki/Descartes%27_theorem and in particular the Special cases. Show inductively that all curvatures remain integers as in http://en.wikipedia.org/wiki/Apollonian_gasket. Now place some well chosen circles into the square and maybe fill up into the corners (as filling between a corner and a circle is not handled with Decartes). All other areas, i.e. in between circles and between circles and square edges you can fill with the Decartes theorem. You can probably show analytically which radii you are going to use up. Now sum their areas.
