Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges

Hint:

$$\sum_{k=1}^{\infty} \frac{\text{ln}\ k}{k^2} < \int_{1}^{\infty}\frac{\text{ln}\ x}{x^2} dx = 1 $$

• Try looking at the maximum value of the function $\displaystyle f(x) = \frac{\ln(x)}{\sqrt{x}}$ on $(1,\infty)$

• $\displaystyle f'(x)= \frac{2-\ln(x)}{2x^{3/2}}$

• $f'(x) <0$ for all $x>e^{2}$. Hence $f(x) \leq f(e^{2})$ for all $x\geq 1$. Hence $\displaystyle \frac{\ln(x)}{\sqrt{x}} \leq \frac{2}{e}$.

• Now $\displaystyle \sum_{k=1}^{\infty} \frac{\ln{(k)}}{k^{2}} = \sum\limits_{k=1}^{\infty} \frac{\ln(k)}{\sqrt{k}} \times \frac{1}{k\sqrt{k}} \leq \frac{2}{e} \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} <\infty$