With $\sigma ( k ) = n+1-k$ we have $$\sum_{k=1}^{n} \frac{k}{(k+\sigma(k))^2} = \sum_{k=1}^n \frac{k}{(n+1)^2} = \frac{n(n+1)}{2(n+1)^2} = \frac{n}{2(n+1)} < \frac{1}{2}. $$

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?

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.