While studying class notes of a senior in number theory, I am unable to deduce this inequality. I can't ask from our instructor as university is closed .
Prove that $$ \sum_{s=2}^n \log^{-2} s < c n \log^{-2} n,$$ where $c$ is a constant.
Please help.
Since OP asked for a clarification of my comment, I am expanding here.
$\log s$ is a monotonically increasing function, $(\log s)^{-2}$ is monotonically decreasing for $s > 1$. This means that in the sum of positive numbers \begin{equation} \sum_{s=2}^n (\log s)^{-2} \, , \end{equation} with $n \geq 2$, each term contributes less than the one before. Hence, we can bound this sum with \begin{equation} \sum_{s=2}^n (\log n)^{-2} \leq \sum_{s=2}^n (\log s)^{-2} \leq \sum_{s=2}^n (\log 2)^{-2} \, , \end{equation} or, in in other words, \begin{equation} (n - 1)(\log n)^{-2} \leq \sum_{s=2}^n (\log s)^{-2} \leq (n - 1)(\log 2)^{-2} \, . \end{equation} Dividing by $n (\log n)^{-2}$, we get \begin{equation} \frac{n - 1}{n} \leq f(n) \leq \frac{n - 1}{n} (\log n)^{2} (\log 2)^{-2} \, , \end{equation} where I have introduced $f(n)$ given by \begin{equation} f(n) = \frac{1}{n (\log n)^{-2}} \sum_{s=2}^n (\log s)^{-2} \, . \end{equation} Taking $n \rightarrow \infty$ in the previous inequality we also find \begin{equation} 1 \leq \lim_{n \rightarrow \infty} f(n) \, . \end{equation} And, since $n \geq 2$, the inequality also gives us \begin{equation} \frac{1}{2} \leq f(n) \, . \end{equation} The current problem we are exploring is, in some measure, whether $f(n)$ has also a finite upper bound. This is, a constant $c'$ that satisfies \begin{equation} f(n) \leq c' \, , \end{equation} or, in other words, \begin{equation} \sum_{s=2}^n (\log s)^{-2} \leq c' n (\log n)^{-2} \, . \end{equation} It helps to look at the plot of $f(n)$:
As I mentioned in the comment, $f(n)$ has a maximum at $f(21) \approx 2.867$ and then seems to decrease, always constrained by $\frac{n-1}{n} \leq f(n)$. Hence, if we can prove that $f(21)$ is a global maximum, then we can take $c' = f(21)$ as the upper bound we so desire.