Integral Test — Question 5

*Question 5 Consider the nested-logarithm series $$\sum_{n=3}^{\infty}\frac1{n\ln n\,\ln(\ln n)}.$$

1. Explain why starting at $n=3$ makes the terms positive and well defined.

2. Verify the Integral Test hypotheses for the corresponding function.

3. Evaluate the improper integral using two successive logarithmic substitutions.

4. State the verdict and identify the slow growth expression responsible for it.

*Question 5 – Solution

Start at $n=3$ so that $\ln n>1$ and $\ln(\ln n)>0$. Let $$f(x)=\frac1{x\ln x\ln(\ln x)}\quad(x\ge3).$$ The denominator is positive, continuous, and strictly increasing, so $f$ is positive, continuous, and decreasing.

First let $u=\ln x$, $du=dx/x$; then let $v=\ln u$, $dv=du/u$: $$\int_3^b\frac{dx}{x\ln x\ln(\ln x)} =\int_{\ln3}^{\ln b}\frac{du}{u\ln u} =\left[\ln(\ln u)\right]_{\ln3}^{\ln b}.$$ Thus the antiderivative grows as $\ln(\ln(\ln b))$, which tends to infinity.

By the Integral Test, $$\boxed{\sum_{n=3}^{\infty}\frac1{n\ln n\ln(\ln n)}\text{ diverges}.}$$ This is another borderline series: three nested logarithms still grow without bound.