Help needed in dedducing an inequality related to logarithms 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. 
 A: 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.
