TITLE: An elementary proof for a series involving logarithm
QUESTION [8 upvotes]: 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?

REPLY [10 votes]: 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 $\frac{111}{160}<\frac{8}{11}.$