This is a follow up question from my previous post: Show $\sum_{k = 1} ^n \frac{(k - 1)n^2}{n(n - k + 1)^2} = O(n^2)$. I now understand the sum on the left is $O(n^2)$. Let me call the sum $f(n)$ for convenience.
It is now claimed in the proofs that I am reading claims $\frac{f(n)}{n^2 \log^2 n} = o(\frac{1}{\log^2 n})$. However, I am not sure if I see this is true
This is essentially saying $$ \lim_n \frac{f(n)}{n^2 \log^2 n} \log^2 n = \lim_n \frac{f(n)}{n^2} = 0. $$ The only information that we have is $\frac{f(n)}{n^2}$ is upper bounded when $n \to \infty$. How could we conclude the claim?
f(n) is between n^2 and pi^2/6 n^2, so your function is between 1 / log^2 n and (pi^2/6) / log^2 n. So it’s not little-o but both big-O and $\theta (1 / log^2 n)$