$ \sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$ Let $ f: \mathbb N  \to \mathbb N$ be a bijective function such that:
$$ \sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$$
Now my question is does any such $f$ exists?
 A: Let $f$ be any given bijection from $\Bbb N\to\Bbb N$. Consider the partial sum $S_n=\sum_{k=1}^n\frac{f(k)}{k^2}$. Suppose $\{t_i\}_{i=1}^n$ is a strictly increasing sequence such that $t_i\in f(\{1,2,...,n\})$. Clearly, $t_i\ge i$. Then, from the rearrangement inequality$$S_n\ge\sum_{i=1}^n\frac{t_i}{i^2}\ge\sum_{i=1}^n\frac i{i^2}$$
Thus, $\lim_{n\to\infty}S_n\ge\lim_{n\to\infty}\sum_{i=1}^n\frac i{i^2}\to\infty$.
A: Define $$g(m)=\sum_{n=1}^m\dfrac{f(n)}{n^2}$$
Now note that if the $f$-image of the set $[m]=\{1,2,\dots, m\}$ is $[m]$ itself, then the minimum value of $g(m)$ is taken when $f(k)=k\:\forall\:1\le k\le m$ (by rearrangement inequality).
Otherwise there exists a subset $S$ of $[m]$ such that $f(S)\subset[m]$ and $f([m]\setminus S)\cap[m]=\emptyset$. In this case, we may note that $g(m)$ is term-wise more than or equal to the partial sum obtained by replacing $f([m]\setminus S)$ by $[m]\setminus f(S)$ and if we do that, the lower bound obtained above using rearrangement inequality again applies. Therefore, we have,
\begin{equation}
\sum_{n=1}^\infty\dfrac{f(n)}{n^2} =\lim_{m\to\infty}g(m)\ge\lim_{m\to\infty}\sum_{i=1}^m\dfrac1n=\infty
\end{equation}
Hence, this series is divergent.
