Is there a bijection of the natural numbers which swaps $\frac{1}{n}$-summable subsets with $\frac{1}{\sqrt{n}}$-summable subsets?

In what follows, I assume that by $\mathbb{N}$ you mean $\mathbb{Z}_{>0}$, this way I don't have to deal with $\frac 1 0$. Of course, the answer is the same for $\mathbb{Z}_{\geq 0}$.

No, such a bijection doesn't exist.

Consider the sequence $(b_n)_{n=1}^\infty$ given by $b_1 = 1$, $b_n = n [\log_2 n]^2$ for $n>1$, where $[x]$ denotes the integer part of $x$. Then the following 2 statements are true:

(1) $\sum_{n=1}^\infty \frac{1}{b_n}$ converges.

(2) Denote $s_k := \sum_{i=1}^k b_i$. Then $\sum_{n=1}^\infty \frac{1}{\sqrt{s_n}}$ diverges.

(1) follows from the fact that for $k\geq 1$, $$ \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{b_n} < \frac{2^k}{b_{2^k}} = \frac{1}{k^2}, $$ and $\sum \frac{1}{k^2}$ converges. To prove (2), note that $s_{2^{k+1}-1} < k^2 \sum_{i=1}^{2^{k+1}-1} i < k^2 2^{2(k+1)}$, so $$ \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{\sqrt{s_n}} > \frac{2^k}{\sqrt{s_{2^{k+1}-1}}} > \frac{1}{2k}, $$ and $\sum \frac{1}{2k}$ diverges.

Now let $f$ be a bijection from $\mathbb{N}$ to $\mathbb{N}$. Set $u_n = \max\{f^{-1}(i)\mid s_{n-1}< i \leq s_n\}$ and $A = \{u_n\mid n\in \mathbb{Z}_{>0}\}$ (assuming $s_0 = 0$). Note that $u_m\neq u_n$ for $m\neq n$ since $f(u_m)\neq f(u_n)$. Also, $u_n\geq b_n$ since $u_n$ is a maximum of $b_n$ positive integers. Then $\sum_A \frac{1}{n} = \sum_{n=1}^\infty \frac{1}{u_n} \leq \sum_{n=1}^\infty \frac{1}{b_n}$, which converges. On the other hand, $f(u_n) \leq s_n$, so $\sum_{f(A)} \frac{1}{\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{\sqrt{f(u_n)}} \geq \sum_{n=1}^\infty \frac{1}{\sqrt{s_n}}$, which diverges.

Edit: Actually, a much more general statement can be proven:

Let $(a_n)_{n\in\mathbb{N}}$ and $(a'_n)_{n\in\mathbb{N}}$ be two sequences of positive numbers. Assume that $\lim\limits_{n\to\infty} a'_n = \lim\limits_{n\to\infty} \frac{a'_n}{a_n} = 0$ and that $\sum_{n\in \mathbb{N}} a_n$ diverges. Then $\left(\mathbb{N}, a_n\right)$ and $\left(\mathbb{N}, a'_n\right)$ are not homeomorphic, i.e., there is no bijection $f: \mathbb{N}\to \mathbb{N}$ such that for every $A\subseteq\mathbb{N}$, $\sum_A a_n$ converges iff $\sum_{f(A)} a'_n$ converges.

Proof: First of all, it's enough to prove the statement for the cases where the sequence $a'_n$ is non-increasing. Since $a'_n>0$ and $a'_n\to 0$, the sequence $a'_n$ can be rearranged in a non-increasing order, i.e., there is a bijection $p:\mathbb{N}\to \mathbb{N}$ such that $a'_{p(n+1)}\leq a'_{p(n)}$ for all $n\in \mathbb{N}$. The sequences $a_{p(n)}$ and $a'_{p(n)}$ satisfy the conditions of the statement, and $f$ is a homeomorphism between $\left(\mathbb{N}, a_{p(n)}\right)$ and $\left(\mathbb{N}, a'_{p(n)}\right)$ iff $p^{-1}\circ f\circ p$ is a homeomorphism between $\left(\mathbb{N}, a_n\right)$ and $\left(\mathbb{N}, a'_n\right)$. So in what follows I assume that $a'_n$ is non-increasing.

Let us denote partial sums of $\sum a_n$ and $\sum a'_n$ by $S_n$ and $S'_n$, i.e., $S_n:= \sum_{k\leq n} a_n$ and $S'_n:= \sum_{k\leq n} a'_n$.

Lemma 1: $\lim\limits_{n\to\infty} \frac{S'_n}{S_n} = 0$.

Proof of Lemma 1: For any $\epsilon>0$, there is $N\in \mathbb{N}$ such that $\frac{a'_n}{a_n}<\frac{\epsilon}2$ for all $n>N$. Then, for $n>N$, $$ \frac{S'_n}{S_n} = \frac{S'_N + \sum\limits_{N<k\leq n} a'_k}{S_n} < \frac{S'_N + \frac{\epsilon}{2}\sum\limits_{N<k\leq n} a_k}{S_n} = \frac{S'_N + \frac{\epsilon}{2}(S_n-S_N)}{S_n}, $$ which is $<\epsilon$ if $S_n> \frac{2}{\epsilon}S'_N - S_N$. Since $\sum a_n$ diverges, $S_n\to\infty$, so $\frac{S'_n}{S_n}<\epsilon$ for all large enough $n$. End of proof of Lemma 1

Now let $f: \mathbb{N}\to \mathbb{N}$ be a bijection. For $\epsilon>0$, denote $M^f_\epsilon := \{n\in\mathbb{N}\mid a'_{f(n)}<\epsilon a_n\}$.

Lemma 2: For all $\epsilon>0$, $\sum_{M^f_\epsilon} a_n$ diverges.

Proof of Lemma 2: Assume that $\sum_{M^f_\epsilon} a_n$ converges, and denote its value by $\mu$.

Since $f$ is a bijection and $a'_n$ is non-increasing, $\sum_{k\leq n} a'_{f(k)} \leq \sum_{k\leq n} a'_{k} = S'_n$, so $$ \frac{S'_n}{S_n} \geq \frac{\sum\limits_{k\leq n} a'_{f(k)}}{S_n} \geq \frac{\sum\limits_{k\leq n,~ k\not\in M^f_\epsilon} a'_{f(k)}}{S_n} \geq \frac{\epsilon\sum\limits_{k\leq n,~ k\not\in M^f_\epsilon} a_{k}}{S_n} \geq \frac{\epsilon(S_n - \mu)}{S_n}. $$ The last expression is $\geq \epsilon/2$ if $S_n>2\mu$; since $S_n\to\infty$, it contradicts Lemma 1. End of proof of Lemma 2

Following your terminology, I will call a subset $X\subseteq \mathbb{N}$ $a$-large if $\sum_X a_n$ diverges. So $M^f_\epsilon$ is $a$-large.

Finally, let us construct a counterexaple set $A$ as a union of pairwise disjoint finite sets $A_m$, $m\in\mathbb{N}$. We construct $A_m$ inductively as follows: consider the set $$ M^f_{2^{-m}} \backslash \left(\{n\in\mathbb{N}\mid a'_{f(n)}\geq 2^{-m}\}\cup\bigcup\limits_{k<m} A_k\right). $$ This set is $a$-large, since it's a difference of $a$-large set $M^f_{2^{-m}}$ and a finite set. (The set $\{n\in\mathbb{N}\mid a'_{f(n)}\geq 2^{-m}\}$ is finite since $a'_n\to 0$ and f is a bijection, and $\cup_{k<m} A_k$ is finite since all $A_k$ are finite.) Thus, it contains a finite subset $X$ such that $\sum_X a_n>1$; let $A_m$ be a minimal such subset w.r.t. inclusion and let $x$ be some element of $A_m$. Then $$ \sum_{f(A_m)} a'_n = \sum\limits_{k\in A_m} a'_{f(k)} = a'_{f(x)} + \sum\limits_{k\in A_m\backslash \{x\}} a'_{f(k)} < 2^{-m} + 2^{-m}\sum\limits_{k\in A_m\backslash \{x\}} a_k \leq 2^{-m+1}. $$ Since $A_m$ are by construction pairwise disjoint, for $A = \cup_{m\in\mathbb{N}} A_m$, $\sum_A a_n = \sum_{m\in\mathbb{N}}\sum_{A_m} a_n$ diverges (by construction, $\sum_{A_m} a_n>1$), and $\sum_{f(A)} a_n = \sum_{m\in\mathbb{N}}\sum_{f(A_m)} a_n < \sum_{m\in\mathbb{N}} 2^{-m+1}$, which converges. End of proof


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.

Tags:

General Topology

Sequences And Series

Real Analysis

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