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

• Now that this question is on the front page again, could someone edit the title so it says "Does there exist a bijective $f:\mathbb{N}\rightarrow\mathbb{N}$..."?
– user856
Oct 7 '10 at 18:39
• Here's a link to the problem and the solution Chandru1 posted below: imc-math.org.uk/imc1999/prob_sol1.pdf Nov 2 '10 at 1:05

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.

• 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? Jan 13 '18 at 10:58
• @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 Jul 21 '21 at 11:30
• (+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.
– robjohn
Jul 21 '21 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.