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Examine if there exists a \(\displaystyle{1-1}\) function \(\displaystyle{f:\mathbb{N}\to \mathbb{N}}\)

such that \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{f(n)}{n^2}<\infty}\)

We show that the series $\sum_{n=1}^\infty\,f(n)/n^2$ diverges for any bijective from $\mathbb{N}$ to $\mathbb{N}$. To see this, since $f$ is a permutation of $\mathbb{N}$, it follows that
$$f(1) + f(2) + \cdots + f(n) \geq 1 + 2 + \cdots + n = \frac{n(n+1)}{2}.$$
Thus, by Abel's summation formula, we have
\begin{eqnarray*}
\sum_{n=1}^N\,\frac{f(n)}{n^2} & = & \sum_{n=1}^{N-1}\,(f(1) + \cdots + f(n)) \left(\frac{1}{n^2}- \frac{1}{(n+1)^2}\right) + \frac{1}{N^2}\sum_{n=1}^N\,f(n)\\
& \geq & \sum_{n=1}^{N-1}\,\frac{n(n-1)}{2}\cdot \frac{2n+1}{n^2(n+1)^2} + \frac{N+1}{2N}\\
& = & \sum_{n=1}^{N-1}\,\frac{(n-1)(2n+1)}{n(n+1)^2} + \frac{N+1}{2N}\\
& \geq & \sum_{n=1}^{N-1}\,\frac{(n-1)}{n(n+1)} + \frac{1}{2}.
\end{eqnarray*}
This implies that $\sum_{n=1}^\infty\,f(n)/n^2$ diverges because $\sum_{n=1}^\infty\,\frac{(n-1)}{n(n+1)}$ diverges.

An alternative proof is based on the Cauchy criterion. Observe that, of $\{f(N+1), f(N+2), \cdots, f(3N)\}$, only $N$ of them are possibly at most $N$, the rest of them must be strictly greater than $N$. This yields the estimate
$$\sum_{n=N+1}^{3N}\,\frac{f(n)}{n^2} \geq \frac{1}{(3N)^2}\,\sum_{n=N+1}^{3N}\,f(n) > \frac{1}{9N^2}\cdot N \cdot N = \frac{1}{9}.$$
It follows that hat $\sum_{n=1}^\infty\,f(n)/n^2$ diverges again by the Cauchy criterion.

mathofusva wrote:We show that the series $\sum_{n=1}^\infty\,f(n)/n^2$ diverges for any bijective from $\mathbb{N}$ to $\mathbb{N}$. To see this, since $f$ is a permutation of $\mathbb{N}$, it follows that
$$f(1) + f(2) + \cdots + f(n) \geq 1 + 2 + \cdots + n = \frac{n(n+1)}{2}.$$
Thus, by Abel's summation formula, we have
\begin{eqnarray*}
\sum_{n=1}^N\,\frac{f(n)}{n^2} & = & \sum_{n=1}^{N-1}\,(f(1) + \cdots + f(n)) \left(\frac{1}{n^2}- \frac{1}{(n+1)^2}\right) + \frac{1}{N^2}\sum_{n=1}^N\,f(n)\\
& \geq & \sum_{n=1}^{N-1}\,\frac{n(n-1)}{2}\cdot \frac{2n+1}{n^2(n+1)^2} + \frac{N+1}{2N}\\
& = & \sum_{n=1}^{N-1}\,\frac{(n-1)(2n+1)}{n(n+1)^2} + \frac{N+1}{2N}\\
& \geq & \sum_{n=1}^{N-1}\,\frac{(n-1)}{n(n+1)} + \frac{1}{2}.
\end{eqnarray*}
This implies that $\sum_{n=1}^\infty\,f(n)/n^2$ diverges because $\sum_{n=1}^\infty\,\frac{(n-1)}{n(n+1)}$ diverges.

Excellent executed mathofusva. I missed the key observation

mathofusva wrote:To see this, since $f$ is a permutation of $\mathbb{N}$, it follows that
$$f(1) + f(2) + \cdots + f(n) \geq 1 + 2 + \cdots + n = \frac{n(n+1)}{2}.$$

and then I gave up. Thank you for this nice solution.