Is there a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the series $\sum_1 ^\infty \frac{1}{n+f(n)} $ is convergent?

I could not solve this. I tried to proceed in following lines:

1) Tried to provide a contradiction:

First let $n \sim m$ iff $\exists k \in \mathbb{Z}$ such that $f^k(n)=m$. This is an equivalence relation. Consider the orbits. For the finite orbits we can compare the series to $\sum_1^\infty \frac{1}{n+n}$. But then I could not figure out how to proceed for infinite orbits.

2) Tried to prove that there is some function:

Let $\{k_n\}$ be a subsequence of $\mathbb{N}$ such that $\sum_0^\infty \frac{1}{k_n}$ converges. Set $f(n)=k_{n}, \forall n \in \mathbb{N}\setminus\{k_n\}$. Then images of each $n$ which are not in the subsequence $k_n$ is defined. Now we have to define images of each $k_n$. Define $f(k_n)=n$ $ \forall n \in \mathbb{N}\setminus \{k_n\}.$.

Could not proceed further. I think My second attempt was going in right direction. My plan was use the fact that all elements here are positive and to construct the function $f$ in such a manner that $\forall n\in \mathbb{N}$ either $n$ or $f(n)$ is in $\{k_n\}$.

  • 3
    $\begingroup$ Decompose $\mathbb N$ into the disjoint union of $S=\{k^2\mid k\in\mathbb N\}$ and $R=\mathbb N\setminus S$. Decompose $S$ into the disjoint union $\bigcup\limits_{k\in\mathbb N}S_k$, where the size of $S_k$ is $2k$ and every element of $S_k$ is smaller than every element of $S_{k+1}$, for every $k$. Then enumerate $R=\{r_k\mid k\in\mathbb N\}$ with $r_k<r_{k+1}$ for every $k$, reorder $\mathbb N$ as $$S_1\ r_1\ S_2\ r_2\ S_3\ \ldots\ S_k\ r_k\ S_{k+1}\ \ldots$$ and consider the corresponding function $f$. Then $$\sum_{n\in S}\frac1{n+f(n)}<\sum_{n\in S}\frac1{n}=\sum_k\frac1{k^2}$$ $\endgroup$
    – Did
    Feb 17 '17 at 12:51
  • 1
    $\begingroup$ ... converges and $r_k\geqslant k^2$ for each $k$, hence $$\sum_{n\in R}\frac1{n+f(n)}<\sum_k\frac1{r_k}\leqslant\sum_k\frac1{k^2}$$ converges hence the whole series converges. $\endgroup$
    – Did
    Feb 17 '17 at 12:51
  • $\begingroup$ @Did, I think you want to order $\mathbb{N}$ as $r_1,S_1,r_2,S_2,r_3,S_3,\ldots$ (providing that $0\notin\mathbb{N}$). But well, if $0\in \mathbb{N}$, then we have to modify this solution only slightly. I would give you an upvote if you posted this as a solution. $\endgroup$ Feb 17 '17 at 13:03
  • $\begingroup$ @Batominovski "I think you want to order N as r1,S1,r2,S2,r3,S3,…" Hmmm, why? What difference does it make? $\endgroup$
    – Did
    Feb 17 '17 at 19:36
  • $\begingroup$ @Did The difference is that $f^{-1}\left(r_k\right)\geq k^2$ may not hold. You have $f^{-1}\left(r_k\right)\geq k^2-1$ instead. (However, nothing really changes.) $\endgroup$ Feb 18 '17 at 2:26

Originally, I gave this example:

$$f (n)=\begin{cases}k,&\ \text { if } n=3^k , \text { with $k $ not a power of $2$}\\ 2^{n-1} , &\ \text { otherwise }\end{cases}$$ (the idea is to push the small numbers further and further down the road so that when they appear they are compensated by the $n $). Then $$ \sum_n\frac1 {n+f (n)}<\sum_{k}\frac 1 {3^k+k}+\sum_n\frac1 {n+2^{n-1}}<\infty. $$

And it is the right idea, but the problem is that such $f$ is not onto. For instance, $2^{26}$ is not in the range of $f$, because when $n=27$, we are using the other branch of $f$ to get $3$.

So we need to tweak the example slightly. Let $$ T=\{3^k:\ k\ \text{ is not a power of } 2\}=\{3^3,3^5,3^6,3^7,3^9,\ldots\} $$ and $$ S=\mathbb N\setminus T=\{1,\ldots,7,8,10,11,\ldots\}. $$ Write them as an ordered sequence, $T=\{t_1,t_2,\ldots\}$ and $S=\{s_1,s_2,\ldots\}$. Now define $$ f(n)=\begin{cases} \log_3 n,&\ \text{ if }\ n=t_k\\ \ \\ 2^{k-1},&\ \text{ if }\ n=s_k \end{cases} $$ One can check explicitly that $$ g(m)=\begin{cases} 3^m,&\ \text{ if $m$ is not a power of $2$}\ \\ \ \\ s_{k+1},&\ \text{ if }\ m=2^k \end{cases} $$ is an inverse for $f$.


Not a new solution, just writing to clarify for myself how the solution of @Martin Argerami: works.

We want $\sum_{n\in \mathbb{N}} \frac{1}{n + f(n)} < \infty$. Consider a covering $\mathbb{N} = A\cup B$. It's enough to have $\sum_{n\in A} \frac{1}{n + f(n)} $, $\sum_{n \in B} \frac{1}{n + f(n)}< \infty$. So it's enoough to find $A$, $B$ so that $$\sum_{n \in A} \frac{1}{f(n)}= \sum_{n \in f(A)} \frac{1}{n} < \infty \\ \sum_{n \in B} \frac{1}{n} < \infty$$

So it's enough to have $B$ so that $\sum_{n\in B} \frac{1}{n} < \infty$ and $f$ mapping $A = \mathbb{N} \backslash B$ to $B$. There are many possibilities here.

  • $\begingroup$ I would not say that "this is how the other solution works" because the other solution relies heavily on infinite orbits while a simple involution between the powers of two and the non-powers of two works fine. $\endgroup$
    – Phira
    Nov 19 '18 at 10:58

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