Packing disjoint family of discs with radii $\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4},\ldots$ inside the unit disc Does there exist a family of discs $\lbrace D_{n}\rbrace_{n=1}^{\infty}$ in the Euclidean plane such that


*

*the radius of $D_{n}$ is $\frac{1}{n+1}$,

*each $D_{n}$ is contained in the unit disc, and  

*$D_{n}\cap D_{m} = \emptyset$ for each $n\neq m$ ?


(I'm not sure what tags are appropriate for this kind question, so if You have any suggestions, You're welcome to inform me about it via comments)
 A: The following is the rigorous construction of desired packing.
Consider the following picture:

On this picture discs with curvatures 2,2,3,3,6,6,6,6,11,11,11,11 are packed inside a unit disc. One can prove correctness of this image by solving quadratic equations. I want to cut circles with curvatures 2,3,4,5,6,... form these: use already obtained circles with curvatures 2 and 3. Use the second circle with curvature 2 for cutting 4,6,8,10,16,... (all even curvatures, starting with 4) using repeating the same procedure scaled to the circle with curvature 2. Use the second 3 for cutting 5 and 9. Use 6 for cutting 7. Use 11 for 11.
Now we have circles with radii 6, 6 and 6. Use the same procedure to obtain circles with curvatures 6*2, 6*3, 6*4, ... = 12, 18, 24, ... from them, each repeating 3 times. Use 12, 12 and 12 for 13, 15, 17; 18 for 19, 21, 23; 24 for 25, 27, 29 and so on.
One can check, that words "use the same procedure" are ok, because if we will use the scheme, described above to cut circles one by one (2, then 3, then 4, then 5, then 6, then 7 in this order) immediately repeating the steps in described smaller circles, we will never use result of the step before the step itself.
