Let $n$ be a positive integer. Show that there exists a one-to-one function $\sigma : \{1,2,...,n\} \to \{1,2,...,n\}$ such that $$\sum_{k=1}^{n} \frac{k}{(k+\sigma(k))^2} < \frac{1}{2}.$$
Problem
Source: PMO
Tags: algebra, number theory, PMO
19.03.2021 12:58
With $\sigma ( k ) = n+1-k$ we have $$\sum_{k=1}^{n} \frac{k}{(k+\sigma(k))^2} = \sum_{k=1}^n \frac{k}{(n+1)^2} = \frac{n(n+1)}{2(n+1)^2} = \frac{n}{2(n+1)} < \frac{1}{2}. $$
20.03.2021 16:30
Gryphos wrote: With $\sigma ( k ) = n+1-k$ we have $$\sum_{k=1}^{n} \frac{k}{(k+\sigma(k))^2} = \sum_{k=1}^n \frac{k}{(n+1)^2} = \frac{n(n+1)}{2(n+1)^2} = \frac{n}{2(n+1)} < \frac{1}{2}. $$ How did u get the motivation for n+1-k?
23.03.2021 15:58
a_bc wrote: Gryphos wrote: With $\sigma ( k ) = n+1-k$ we have $$\sum_{k=1}^{n} \frac{k}{(k+\sigma(k))^2} = \sum_{k=1}^n \frac{k}{(n+1)^2} = \frac{n(n+1)}{2(n+1)^2} = \frac{n}{2(n+1)} < \frac{1}{2}. $$ How did u get the motivation for n+1-k? Intuition says that that the first denominators should be large and the last should be as small as possible. Also this could be verified by brute force calculations.