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I started by taking the example $f(n) = n$ which is the identity bijection and $\sum{\frac{f(n)}{n^2}} = \sum{\frac{1}{n}} > \infty $

Suppose we change the values of the bijection at finitely many places(say , $i_1,i_2,\cdots i_k$) then we would still have the series $\sum_{i=i_{k+1}}^{\infty}{\frac{f(i)}{i^2}} = \sum_{i=i_k}^{\infty}\frac{1}{i_k}$ converging to $\infty$

Suppose we pick up a bijection in such a way so that $f(n) < n$ at infinitely many places then there are infinitely many places at which $f(n) >n$ then we can pick up a subsequence say $(n_k)$ such that $f(n_k) > n_k , \forall k \ge 1$ . Then $\sum_{i = n_k}^{\infty}\frac{f(n_k)}{n_k^2} > \sum_{i = n_k}^{\infty} \frac{1}{n_k} > \infty $. So from here I tried to conclude that $\sum\frac{f(n)}{n^2}$ converges to $\infty$.

Hence from here I tired to conclude that there doesnot exist any bijection $f(n)$ such that the $\sum{\frac{f(n)}{n^2}}$ converges .

How do I fix my reasoning?

Antimony
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    I don't follow the argument of your third paragraph. It could be that $n_k=k^2$ and then $\sum\frac{1}{n_k}$ converges. – ancient mathematician Aug 03 '21 at 06:32
  • $f(n_k) > n_k$ then $\frac{f(n_k)}{(n_k)^2} > \frac{1}{n_k}$. Is the series $\sum{\frac{1}{n_k}}$ convergent?(I am not sure though, but by cauchy's condensation test it shouldnt be so ) – Antimony Aug 03 '21 at 06:38
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    @Antimony: By $\sum\frac 1{n_k}$, would you mean $\sum_{k\ge 1}\frac 1{n_k}$, or $\sum_{r\ge 0}\frac 1{n_k+r}$? The first can converge with $n_k=k^a$ with $\Re(a)\gt 1$ to $\zeta(a)$ (being an upper bound suffices to show convergence, though); the second one is a tail of the harmonic series, hence diverges, no matter the choice of $n_k$. What precisely is the role of the index $i$ in $\sum_{i=n_k}^\infty\frac 1{n_k}$? – Prasun Biswas Aug 03 '21 at 06:42
  • Also: https://math.stackexchange.com/q/4030456/42969, https://math.stackexchange.com/q/4059185/42969 – all found with Approach0 – Martin R Aug 03 '21 at 07:02

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There is indeed no bijection such that the series is convergent. If you want to be more rigorous, we can prove the result using finite sums. Let $f : \mathbb{N} \rightarrow \mathbb{N}$ a bijection, $n \in \mathbb{N}$, and $a_k=f(k)$ for $k \in [|1,n|]$. Now, we note $(b_k)_{1 \leq k \leq n}$ the permutation of $[|1,n|]$ that is ordered in the same order as $(a_k)_{1 \leq k \leq n}$, i.e. if $a_i>a_j$, we have $b_i>b_j$. We therefore have: $$\sum_{k=1}^n \frac{a_k}{k^2} \geq \sum_{k=1}^n \frac{b_k}{k^2}$$ Since $b_1 \leq … \leq b_n$ and $1 \geq \frac{1}{2} \geq … \geq \frac{1}{n^2}$, we can now use the rearrangement inequality (https://en.m.wikipedia.org/wiki/Rearrangement_inequality) to say that: $$\sum_{k=1}^n \frac{b_k}{k^2} \geq \sum_{k=0}^n \frac{k}{k^2}$$ $$\Rightarrow \sum_{k=1}^n \frac{f(k)}{k^2} \geq \sum_{k=0}^n \frac{1}{k}$$ Since $RHS$ diverges as $n \rightarrow +\infty$, $LHS$ does too, and the series is divergent.

Jujustum
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