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

  • $\begingroup$ Now that this question is on the front page again, could someone edit the title so it says "Does there exist a bijective $f:\mathbb{N}\rightarrow\mathbb{N}$..."? $\endgroup$
    – user856
    Oct 7, 2010 at 18:39
  • 4
    $\begingroup$ Here's a link to the problem and the solution Chandru1 posted below: imc-math.org.uk/imc1999/prob_sol1.pdf $\endgroup$ Nov 2, 2010 at 1:05

4 Answers 4


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.

  • $\begingroup$ I don't understand how you draw the conclusion from the fact, that swapping reduces the value of the sum. Can you add a bit more detail? $\endgroup$
    – philmcole
    Jan 13, 2018 at 10:58
  • $\begingroup$ @philmcole the swapping argument is that (if finite) $\sum\frac{f(n)}{n^2}$ would be minimised by swapping only when $f(n)$ is an increasing function, as it is the only case where no more swapping would be possible, and $f(n)=n$ is the only increasing bijective function $\endgroup$
    – Henry
    Jul 21, 2021 at 11:30
  • 2
    $\begingroup$ (+1) The Rearrangement Inequality says that $\sum\limits_{n=1}^\infty\frac{f(n)}{n^2}$ is minimized by $f(n)=n$. That is because $\frac1{n^2}$ is decreasing and $f(n)=n$ is increasing. $\endgroup$
    – robjohn
    Jul 21, 2021 at 11:38

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

To prove this we can use Abel transformation.

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

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 $c > 0$.


Since this question has been closed as a duplicate of this one, I will only assume, as in the other question, that $f$ is injective, and show that the series must diverge. In doubt, I will also assume $\mathbb{N} = \{0, 1, 2, \dots\}$.

Let $S_N = \sum_{n=1}^N \frac{f(n)}{n^2}$. The sum $S_N$ assigns nonnegative "weights" $f(1)$, $f(2)$,... , $f(N)$, to the numbers $\frac{1}{1^2}$, $\frac{1}{2^2}$, ....,$\frac{1}{N^2}$, in that order, and takes the sum.

Now say we would like to make the sum defining $S_N$ as small as possible by rearranging the weights. It is clear that to do this the largest weights must be assigned to the smallest numbers. Therefore if the weights $f(1)$, $f(2)$, ..., $f(N)$, are arranged in increasing order as $a_1$, $a_2$, ..., $a_N$, we must have the inequality $$S_N \geq \sum_{n=1}^N \frac{a_n}{n^2}.$$

Because $f$ is assumed injective, it is easy to show by induction that $a_n \geq n - 1$. Thus the partial sums $S_N$ satisfy the inequality $$S_N \geq \sum_{n=1}^N \frac{n-1}{n^2}.$$ Therefore the series diverges by comparison with, for instance, $\sum_2^{+\infty} \frac{1}{2n}$


We show that the for any $f: \mathbb{N} \rightarrow \mathbb{N}$ bijective this is not cauchy. Suppose it is for given $\epsilon > 0$ there exists $N$ such that $\sum_{n=N}^{2N} \frac{f(n)}{n^2} < \epsilon$. We have $\sum_{n=N}^{2N} \frac{f(n)}{n^2} \geq \frac{1}{(2N)^2}\sum_{n=N}^{2N}f(n)\geq \frac{1}{(2N)^2} \frac{N(N+1)}{2}=\frac{(N+1)}{8N}=\frac{1}{8}+\frac{1}{8N}$. If we choose $\epsilon < \frac{1}{8}$ we get a contradiction.


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