TITLE: Is there a bijection of the natural numbers which swaps $\frac{1}{n}$-summable subsets with $\frac{1}{\sqrt{n}}$-summable subsets?
QUESTION [16 upvotes]: Let me start with a precise statement of the question.  For a subset $A\subseteq \mathbb{N}$ and a series of positive real numbers $\sum_{n=0}^\infty a_n$, I'll use the notation $\sum_A a_n$ as a shorthand for $\sum_{n\in A} a_n$.

Is there a bijection $f:\mathbb{N}\rightarrow \mathbb{N}$ with the property that for every $A\subseteq \mathbb{N}$, $\sum_A \frac{1}{n}$ converges iff $\sum_{f(A)} \frac{1}{\sqrt{n}}$ converges?

Background:
Fix a series of positive terms $\sum_{n=0}^\infty a_n$.  Given a subset $A\subseteq \mathbb{N}$, call it $a_n$-small if $\sum_A a_n$ converges.  The following proposition is easy to prove:

Proposition:  The set $\{A\subseteq \mathbb{N}: A\text{ is } a_n\text{-small}\}\cup \{\mathbb{N}\}$ is a topology of closed sets on $\mathbb{N}$.

I will use the notation $(\mathbb{N},a_n)$ to refer to this topology.  Then one can ask how topological properties of $(\mathbb{N},a_n)$ are related to series properties of $\sum a_n$.  For example, I can show:

Proposition:  The following are equivalent.

$\sum a_n$ converges.

$(\mathbb{N},a_n)$ is discrete.

$(\mathbb{N},a_n)$ is disconnected.

$(\mathbb{N},a_n)$ is Hausdorff



And there are other nice things.  For example, $(\mathbb{N},a_n)$ is compact iff $(\mathbb{N},a_n)$ is cofinite iff $\liminf a_n > 0$.
With this language, my question can be reformulated as...

Are $\left(\mathbb{N}, \frac{1}{n}\right)$ and $\left(\mathbb{N}, \frac{1}{\sqrt{n}}\right)$ homeomorphic?

I have made very little progress on this.  Of course, the identity function $i:(\mathbb{N}, \frac{1}{\sqrt{n}})\rightarrow (\mathbb{N}, \frac{1}{n})$ is a continuous bijection, but the inverse map is not continuous.  Also, if there is such a bijection, there there is such a bijection with agrees with $i$ on any preassigned finite set.
Edit  I thought I add in a slightly suprising (to me, at least) example when things work out to be homeomorphic.
Begin with a convergent series $\sum c_n$. and divergent series $\sum a_n$ with $\lim a_n = 0$.  Create a new series $b_n$ using all the terms of $c_n$ and $a_n$ (in whatever order you wish).  Then $(\mathbb{N}, a_n)$ and $(\mathbb{N}, b_n)$ are homeomorphic.  The idea is that since $\lim a_n = 0$, there is a convergent infinite subseries $\sum_A a_n$.  Then we use a bijection which with $A$ and $A\cup \{\text{indices of }c_n\}$ to "squeeze" the $c_n$ in without changing the topology.  Of course, I am glossing over many details, but I can include them if desired.

REPLY [4 votes]: I claim that there is no such bijection for any distinct exponents not exceeding $1$.

Lemma If positive vanishing sequences $(a_n),(b_n)$ are such that $\sum_n a_n = \infty$, $b_n = o(a_n)$, $n\to\infty$, then there exists $A\subset \mathbb N$ such that $\sum_A a_n=\infty$, $\sum_A b_n<\infty$. 

Proof Let $N_k$ be such that $b_n<a_n/k$, $n\ge N_k$. Since $(a_n)$ vanishes and $\sum_n a_n = \infty$,  there exist some $n_1'>n_1> N_1$ such that $\sum_{n=n_1}^{n_1'} a_n \in (1,2)$. Inductively, for each $k\ge 2$ there exist  $n_k'>n_k > N_k\vee n_{k-1}'$ such that $\sum_{n=n_k}^{n_k'} a_n \in (1/k,2/k)$. Obviously, $A = \bigcup_{k\ge 1} \{n_k,n_k+1,\dots,n_k'\}$ is as required.

Now assume that for exponents $a<b\le 1$ there exists a bijection $f\colon \mathbb N\to \mathbb N$ such that $\sum_A n^{-b}$ converges iff $\sum_{f(A)} n^{-a}$ converges. 
Choose some $r\in (1,b/a)$. Clearly, $\sum_{n: f(n)>n^{r}} \frac{1}{f(n)} <\infty$ and $\sum_{n} \frac{1}{f(n)} =\infty$, so $\sum_{n: f(n)\le n^{r}} \frac{1}{f(n)} =\infty$. Therefore, $\sum_{n: f(n)\le n^{r}} \frac{1}{f(n)^a} =\infty$. If $f(n)\le n^r$, then $f(n)^a\le n^{ra} = o(n^b)$, $n\to\infty$. Setting $a_n = 1/f(n)^{a}$ and$^*$ $b_n = 1/n^b$ and applying lemma, we get that $\sum_A f(n)^{-a}=\infty$ and $\sum_A n^{-b} <\infty$, a contradiction.

$^*$ apologies for the pun.