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.

  
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*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

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.
