# Does there exist a bijective $f:\mathbb{N} \to \mathbb{N}$ such that $\sum f(n)/n^2$ converges?

We know that $\displaystyle\zeta(2)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and it converges.

• Does there exists a bijective map $f:\mathbb{N} \to \mathbb{N}$ such that the sum $$\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^2}$$ converges.

If our $s=2$ was not fixed, then can we have a function such that $\displaystyle \zeta(s)=\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges

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Now that this question is on the front page again, could someone edit the title so it says "Does there exist a bijective $f:\mathbb{N}\rightarrow\mathbb{N}$..."? – Rahul Narain Oct 7 '10 at 18:39

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– Jonas Meyer Nov 2 '10 at 1:05

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@Chandru1: Please do not post others' problems and solutions as your own. – Jonas Meyer Nov 2 '10 at 1:09

## 2 Answers

For $s>2$ you can take $f(n)=n$.

For $s=2$ if you have $m < n$ and $a=f(m) > b=f(n)$ then $$\frac{a}{m^2}+\frac{b}{n^2}>\frac{b}{m^2}+\frac{a}{n^2}$$ (proved either naively or as a case of the "rearrangement inequality") so the sum $\sum f(n)/n^2$ can be reduced by swapping the values $f(m)$ and $f(n)$. Hence $$\sum\frac{f(n)}{n^2}\ge\sum\frac{n}{n^2}$$ which is divergent.

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For $s=2$ the answer is negative. This series doesn't converge.

To prove this we can use Abel transformation.

$$\sum_{n=1}^{n=N} \frac{f(n)}{n^2} = \sum_{n=1}^{n=N} (\sum_{k=1}^{k=n} f(k)) (\frac{1}{n^2} - \frac{1}{(n + 1)^2}) + (\sum_{n=1}^{N} \frac{1}{n^2})\frac{1}{(N+1)^2}$$

Since $f$ is bijection, $\sum_{k=1}^{k=n} f(k) \ge \frac{n^2}{2}$ hence the first sum is greater than $\sum_{n=1}^{N}\frac{c}{n}$ for some $c > 0$.

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