# Convergence or Divergence using Limits

Use a comparison test to determine whether or not the following improper integrals converge or diverge $$\int_{2}^\infty \frac{1}{\ln(x)}dx.$$ I'm stuck thinking of a function to compare it to.

We have by the L'Hôpital's rule $$\lim_{x\to\infty}\frac{\ln x}{x}=0$$ so for $x$ sufficiently large we have $$\frac1{\ln x}\ge\frac1x$$ so the given integral is divergent.
it obviously diverges because $\frac{1}{\log x } \geq \frac{1}{x}$ which diverges. Compare to harmonic sum/integral.
Set $\ln x=y$ then $dx=e^ydy$ so the integral becomes $$\int_{\ln 2}^\infty \frac{e^y}{y}dy$$