## Calculus (3rd Edition)

Given $$\sum_{m=2}^{\infty}\frac{1}{\ln m}$$ Since for $m\geq2$ \begin{align*} \frac{1}{\ln m} \geq \frac{1}{m} \end{align*} Compare with $\displaystyle\sum_{m=2}^{\infty}\frac{1}{m}$, a divergent $p-$series $(p=1)$; then $\displaystyle\sum_{m=2}^{\infty}\frac{1}{\ln m}$ also diverges.