An elementary proof for a series involving logarithm I want to prove that

$$ \sum_{x=1}^{\infty} \frac{\ln x}{x^2} \leq 1$$

in an elementary way, but haven't found such a proof yet. Would you please share with us if you find any elementary proof of the inequality?

Also, is there any elementary method to find upper bounds for series involving logarithms?
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
$$ \sum_{n\geq 1}\frac{\log n}{n^2} \leq \frac{\log 2}{4}+\frac{\log(4)+1}{3} &lt; 1 \tag{2}$$
since $\log(2)&lt;\frac{8}{11}$. 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 $\frac{111}{160}&lt;\frac{8}{11}.$
Have you tried using a taylor expansion for ln(x), Then dividing this by x^2. It should always be 1/x^n for each term in that expansion.