How to prove $\sum_{k=2}^{n} \frac{\ln^2 k}{k^2}<2-\frac{\ln^2 n}{n}$? If $n\in \mathbb{N}$ and $n\geq 2.$
Prove
$$\sum_{k=2}^{n} \frac{\ln^2 k}{k^2}<2-\frac{\ln^2 n}{n}$$
Maybe there is a simple solution?
 A: Hint: Integration by parts gives
$$
\begin{align}
\int\frac{\log(t)^2}{t^2}\,\mathrm{d}t
&=-\frac{\log(t)^2}{t}+2\int\frac{\log(t)}{t^2}\,\mathrm{d}t\\
&=-\frac{\log(t)^2}{t}-2\frac{\log(t)}t+2\int\frac1{t^2}\,\mathrm{d}t\\
&=C-\frac{\log(t)^2+2\log(t)+2}t
\end{align}
$$
A: This is my attempt at proving this by induction.
Clearly, the inequality is true for $k=2$. Assume that the inequality is true for $k=n-1$ for $n > 2$. We will prove that it is true for $k=n$. Here is the inequality for $k=n-1$:
$$\sum_{k=2}^{n-1} \frac{\ln^2k}{k^2} < 2-\frac{\ln^2(n-1)}{n-1}$$
Here is the inequality for $k=n$:
$$\sum_{k=2}^{n-1} \frac{\ln^2k}{k^2}+\frac{\ln^2n}{n^2} < 2-\frac{\ln^2(n-1)}{n-1}+\frac{\ln^2(n-1)}{n-1}-\frac{\ln^2n}{n}$$
Thus, if we can prove $\frac{\ln^2n}{n^2}<\frac{\ln^2(n-1)}{n-1}-\frac{\ln^2n}{n}$, we can combine it with the inequality from $k=n-1$ to prove the inequality for $k=n$.
Now, this is where I get stuck. The inequality seems to be true after graphing the two functions, but I don't know how to actually prove it. Hopefully, though, this will help you a little bit.
