IMC 1999 · Day 1 · P2

No. For, let $\pi$ be a permutation of $\mathbb{N}$ and let $N \in \mathbb{N}$. We shall argue that $$\sum_{n=N+1}^{3N} \frac{\pi(n)}{n^2} > \frac{1}{9}.$$ In fact, of the $2N$ numbers $\pi(N+1), \dots, \pi(3N)$ only $N$ can be $\le N$ so that at least $N$ of them are $> N$. Hence $$\sum_{n=N+1}^{3N} \frac{\pi(n)}{n^2} \ge \frac{1}{(3N)^2} \sum_{n=N+1}^{3N} \pi(n) > \frac{1}{9N^2} \cdot N \cdot N = \frac{1}{9}.$$

Let $\pi$ be a permutation of $\mathbb{N}$. For any $n \in \mathbb{N}$, the numbers $\pi(1), \dots, \pi(n)$ are distinct positive integers, thus $\pi(1) + \dots + \pi(n) \ge 1 + \dots + n = \frac{n(n+1)}{2}$. By this inequality, $$\begin{align*} \sum_{n=1}^{\infty} \frac{\pi(n)}{n^2} &= \sum_{n=1}^{\infty} \bigl( \pi(1) + \dots + \pi(n) \bigr) \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \ge \\ &\ge \sum_{n=1}^{\infty} \frac{n(n+1)}{2} \cdot \frac{2n+1}{n^2 (n+1)^2} = \sum_{n=1}^{\infty} \frac{2n+1}{2n(n+1)} \ge \sum_{n=1}^{\infty} \frac{1}{n+1} = \infty. \end{align*}$$