Inequality involving sum of logarithms and hidden zeta-function I would like to prove the following estimation: if $n \ge 2$ is a natural number, then
$$\sum_{k=2}^n \frac{\log^2 k}{k^2} <2 - \frac{\log^2 n}{n}.$$
I have noticed that LHS is indeed bounded by proving that
$$\sum_{k = 2}^\infty \frac{\log ^2 k}{k^2} = \zeta''(2) \approx 1.98928$$
and then check with an aid of computer that for $n \ge 7407$ we have $\frac 1 n \log^2 n < 2 - \zeta''(2)$, thus almost furnishing the proof (one also has to verify that the estimation is true for smaller $n$). Is there another way to tackle this problem using (hopefully) only pencil and paper?
 A: We have:
$$ \sum_{k=2}^{n}\frac{\log^2(k)}{k^2}=\zeta''(2)-\sum_{k>n}\frac{\log^2(k)}{k^2}\tag{1}$$
and $f(x)=\frac{\log^2(x)}{x^2}$ is a decreasing function on $[3,+\infty)$, hence:
$$ \sum_{k>n}\frac{\log^2(k)}{k^2}>\int_{n+1}^{+\infty}\frac{\log^2(x)}{x^2}\,dx = \frac{1+(1+\log(n+1))^2}{n+1}>\frac{\log^2(n)}{n}\tag{2}$$
and you inequality can be improved up to:
$$\boxed{\forall n\geq 2,\qquad \sum_{k=2}^{n}\frac{\log^2(k)}{k^2}<\zeta''(2)-\frac{\log^2(n)}{n}.}\tag{3}$$
A: You can also use the Abel's summation. We have $$S=\sum_{k=2}^{n}\frac{\log^{2}\left(k\right)}{k^{2}}=\sum_{k=1}^{n}\frac{\log^{2}\left(k\right)}{k^{2}}=\frac{\log^{2}\left(n\right)}{n}+2\int_{1}^{n}\frac{\left\lfloor t\right\rfloor \log\left(t\right)\left(\log\left(t\right)-1\right)}{t^{3}}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. Now since $t-1\leq\left\lfloor t\right\rfloor \leq t
 $ we have $$\begin{align}S\leq
 & \frac{\log^{2}\left(n\right)}{n}+2\int_{1}^{n}\frac{\log\left(t\right)\left(\log\left(t\right)-1\right)}{t^{2}}dt
  \\ =
  & -\frac{\log^{2}\left(n\right)}{n}-4\frac{\log\left(n\right)}{n}-\frac{2}{n}+2
 \end{align}$$ and $$\begin{align}S\geq
  & \frac{\log^{2}\left(n\right)}{n}+2\int_{1}^{n}\frac{\left(t-1\right)\log\left(t\right)\left(\log\left(t\right)-1\right)}{t^{3}}dt
  \\ =
  & -\frac{\log^{2}\left(n\right)}{n}-2\frac{\log^{2}\left(n\right)}{n^{2}}-4\frac{\log\left(n\right)}{n}-\frac{2}{n}+2
 \end{align}$$ so 

$$\color{blue}{-\frac{\log^{2}\left(n\right)}{n}-2\frac{\log^{2}\left(n\right)}{n^{2}}-4\frac{\log\left(n\right)}{n}-\frac{2}{n}+2}\leq S \leq \color{red}{-\frac{\log^{2}\left(n\right)}{n}-4\frac{\log\left(n\right)}{n}-\frac{2}{n}+2}
 $$ 

and obviously $$S\leq-\frac{\log^{2}\left(n\right)}{n}-4\frac{\log\left(n\right)}{n}-\frac{2}{n}+2\leq2-\frac{\log^{2}\left(n\right)}{n}.$$
