Is there a countable cover of $\mathbb{R}^2$ by balls $B(x_n, n^{-1/2})$? Is it possible to cover all of $\mathbb{R}^2$ using balls $\{ B(x_n,n^{-1/2})\}_{n=1}^\infty$ of decreasing radius $n^{-1/2}$? I know that if we chose e.g. radius $n^{-1}$ it could never work because $\sum \pi (n^{-1})^2 < \infty$. But in this case the balls cover an infinite amount of area, so it seems that it may be possible to construct.
Edit: It can definitely be done with balls of radius $n^{-1/2+\epsilon}$ for any $\epsilon > 0$.
 A: Yes, this can be done.
Let $n\in\mathbb N$, consider $m\in\{1,\dots n\}$, and set $$a(n,m)=n^2+(m-1)n.$$ Note then that $$\sum_{k=a(n,m)+1}^{a(n,m+1)}\frac{1}{\sqrt{k}}\geq\sum_{k=a(n,m)+1}^{a(n,m+1)}\frac{1}{\sqrt{n^2+mn}}=\frac{n}{\sqrt{n^2+mn}}\geq\frac{n}{\sqrt{n^2+n^2}}=\frac{1}{\sqrt{2}}.$$ Therefore, if we consider adjacent squares, this shows that we can cover the strip $\left[0,\frac{1}{\sqrt{2}}\right]\times\left[0,\frac{1}{n\sqrt{2}}\right]$ by squares of the form $$S\left(x_{a(n,m)+1},\frac{1}{\sqrt{a(n,m)+1}}\right),S\left(x_{a(n,m)+2},\frac{1}{\sqrt{a(n,m)+2}}\right),\dots,$$$$S\left(x_{a(n,m+1)},\frac{1}{\sqrt{a(n,m+1)}}\right),$$ for any fixed $m\in\{1,\dots n\}$. If we now repeat this process for the strips $$\left[0,\frac{1}{\sqrt{2}}\right]\times\left[\frac{1}{n\sqrt{2}},\frac{2}{n\sqrt{2}}\right],\,\,\left[0,\frac{1}{\sqrt{2}}\right]\times\left[\frac{2}{n\sqrt{2}},\frac{3}{n\sqrt{2}}\right],\dots\left[0,\frac{1}{\sqrt{2}}\right]\times\left[\frac{n-1}{n\sqrt{2}},\frac{n}{n\sqrt{2}}\right],$$ this shows that we can always cover the square $\left[0,\frac{1}{\sqrt{2}}\right]^2$ by squares of the form $S(x_i,i^{-1/2})$, for $i\in\{n^2,n^2+1,\dots 2n^2\}$.
Repeating this process for all $n\in\mathbb N$ and moving the squares on $\mathbb R^2$, we actually get a covering of all of $\mathbb R^2$ in that way. We then get a covering by balls by circumscribing a ball outside each one of the squares.
A: With disks centered on an $n{+}1\times n{+}1$ grid, we can cover $[0,1]^2$ with $(n+1)^2$ disks of radius $\frac1{n\sqrt2}$.
Therefore, we can cover $[0,1]^2$ with $\left(\left\lceil\sqrt{\frac n2}\,\right\rceil+1\right)^2$ disks of radius $\frac1{\sqrt{n}}$.
For $n\ge294$, $\left(\left\lceil\sqrt{\frac n2}\,\right\rceil+1\right)^2\le\frac23n$. Thus, we have shown

For $n\ge294$, we can cover $[0,1]^2$ with at most $\frac23n$ disks of radius $\frac1{\sqrt{n}}$.

Enumerate the squares $\big\{[i,i+1]\times[j,j+1]:i,j\in\mathbb{Z}\big\}=\big\{S_k:k=1,2,3,\dots\big\}$
Cover $S_k$ with $196\cdot3^{k-1}$ disks with radii from $\frac1{\sqrt{98\cdot3^{k-1}+1}}$ to $\frac1{\sqrt{294\cdot3^{k-1}}}$.
A: It's true for any $r_n$ with $\sum r_n^2=\infty$.
Say $S(x,r)$ is the square with center $x$ and side length $r$.
It's enough to show that if $\sum r_n^2>4$ then there exist $x_n$ such
that $S(x_n,r_n)$ covers $[0,1]^2$.
For every $n$ there exists $j$ with $2^{-j}\le r_n<2^{-j+1}$. Replace $r_n$ with $2^{-j}$:
It's enough to show that if each $r_n$ is of the form $2^{-j}$ and $\sum r_n>1$ then there exist $x_n$ such that $S(x_n,r_n)$ cover the unit square.
A dyadic interval is an interval $[j2^{-n},(j+1)2^{-n}]$. A dyadic square is the product of two dyadic intervals with the same side length.
Start: If there exists $n$ with $r_n\ge1$ we're done. Suppose not. Divide $[0,1]^2$ into four dyadic squares of side length $1/2$. Call them $Q$'s.
If there are four $n$ such that $r_n=1/2$ we're done. Suppose not. Cover as many $Q$'s as possible with $S(x_n,r_n)$ with $r_n=1/2$.
Discard the $r_n$'s we just used. Note we used all the $r_n$ that equal $1/2$. Discard the $Q$'s we just covered. Divide each remaining $Q$ into four subsquares of side length $1/4$.
If there are enough $n$ with $r_n=1/4$ to cover all the reminaing $Q$'s we're done. Suppose not. Cover as many $Q$'s as possible with $S(x_n,r_n)$ where $r_n=1/4$. Note we've used all the $r_n$ that equal $1/4$. Discard those $r_n$, discard the $Q$'s we just covered. Divide each remaining $Q$ into four suubsquares of side length $1/8$.
Etc.
If at some point we're done then we're done. Suppose not. Then we've used up all the $r_n$ covering essentially disjoint subsquares of $[0,1]^2$; hence $\sum r_n^2\le 1$.
