Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n \ge 1$, there is a permutation of $\{1,2,\ldots, n\}$, i.e., some $\pi \in S_n$, such that $$\sum_{k=1}^n \frac{1}{k \cdot \pi(k)} = 1.$$
Here's a question motivated by some related MathOverflow questions of Zhi-Wei Sun.
Show that, for any $n \ge 1$, there is a permutation of $\{1,2,\ldots, n\}$, i.e., some $\pi \in S_n$, such that $$\sum_{k=1}^n \frac{1}{k \cdot \pi(k)} = 1.$$
Copyright © 2021 JogjaFile Inc.
Take $\pi(k)=k+1$ except for $\pi(n)=1$. Then $$\sum_{k=1}^n\frac1{k\pi(k)}=\frac1n+\sum_{k=1}^{n-1}\frac1{k(k+1)} =\frac1n+\sum_{k=1}^{n-1}\left(\frac1k-\frac1{k+1}\right)=1. $$