
> 
> Let $\pi:\mathbb{N}\to\mathbb{N}$ be a bijection. Then does there exist another bijection $\nu:\mathbb{N}\to\mathbb{N}$ and a constant $C$ such that
>  $$
> \frac{1}{n} + \frac{1}{\pi(n)} \leq \frac{C}{\nu(n)}
> $$
>  for all $n$? If so, can the constant be chosen independent of $\pi$?
> 
> 
> 


While the harmonic sequence $(\frac{1}{n})\_{n\in\mathbb{N}}$ is what comes up in my application, I imagine that a good answer will be able to make a much more general statement about a suitable class of sequences. But I'd be perfectly happy with an answer to the question above.



Adding the harmonic sequence and a permutation of it
Yes, we may achieve even $\nu(n)\leqslant 2\min(n,\pi(n))$. Indeed, define $\rho(n)=\min(n,\pi(n))$ and let $t\_1,t\_2,\dots$ be an enumeration of the positive integers such that $\rho(t\_1)\leqslant \rho (t\_2)\leqslant \rho(t\_3)\leqslant \dots$. Then $\rho(t\_k)\geqslant k/2$, otherwise we may find three different numbers with the same values of $\rho$. Therefore defining $\nu(t\_k)=k$ we get $\nu(t\_k)=k\leqslant 2\rho(t\_k)$ as desired.


This results in
$$
\frac{1}{n} + \frac{1}{\pi(n)} \leq \frac{2}{\rho(n)} \leq \frac{4}{\nu(n)},
$$
as was to be shown.


