Inequality involving sum of logarithms and hidden zeta-function

Question

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?

Answer 1

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: $$\boxed{\forall n\geq 2,\qquad \sum_{k=2}^{n}\frac{\log^2(k)}{k^2}<\zeta''(2)-\frac{\log^2(n)}{n}.}\tag{3}$$

Answer 2

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