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? 
 A: 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 
$$\begin{align}\sum_{n=1}^\infty \frac{\ln n}{n^2} &= \frac{\ln 2}{4} + \frac{\ln 3}{9}+\sum_{n=4}^\infty \frac{\ln n}{n^2} \\&< \frac{\ln 2}{4} + \frac{\ln 3}{9}+\int_3^\infty \frac{\ln x}{x^2} \, dx \\&\approx 0.29535  + 0.69954 \\&= 0.99489\end{align}$$
A: 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 $\frac{111}{160}<\frac{8}{11}.$
A: The 4th term and the rest can be replaced by the smaller $x^{-2}$, if we add the area between the graphs for $x^{-2}$ and $\log (x)x^{-2}$ starting one step earlier at $x=3$.
$$\sum_{x=1}^3\log(x)x^{-2}+\sum_{x=4}^{\infty}x^{-2}+\int_3^{\infty}(\log (x)x^{-2}-x^{-2})dx=\frac{6 \pi ^2-49+16 \log (3)+\log (512)}{36}$$
which is less than $1$.
A: 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.
