An elementary proof for a series involving logarithm

An alternative way is to exploit creative telescoping. One may prove the inequality $$ \forall n\geq 2,\qquad \frac{\log n}{n^2}\leq \frac{\log(n+1)+1}{n}-\frac{\log(n+2)+1}{n+1}\tag{1} $$ and deduce from it that $$ \sum_{n\geq 1}\frac{\log n}{n^2} \leq \frac{\log 2}{4}+\frac{\log(4)+1}{3} < 1 \tag{2}$$ since $\log(2)<\frac{8}{11}$. This can be shown via $$ 0\leq \int_{0}^{1}\frac{x^3(1-x)^3}{1+x}\,dx = \frac{111}{20}-8\log(2) \tag{3}$$ and $\frac{111}{160}<\frac{8}{11}.$

Note that

$$\int \frac{\ln x}{x^2} \, dx = -\frac{1+ \ln x}{x}+ C$$

Since $\ln x / x^2$ is decreasing for $x > 2$, we have

$$\begin{align}\sum_{n=1}^\infty \frac{\ln n}{n^2} &= \frac{\ln 2}{4} + \frac{\ln 3}{9}+\sum_{n=4}^\infty \frac{\ln n}{n^2} \\&< \frac{\ln 2}{4} + \frac{\ln 3}{9}+\int_3^\infty \frac{\ln x}{x^2} \, dx \\&\approx 0.29535 + 0.69954 \\&= 0.99489\end{align}$$