Integral Test — Question 10

*Question 10 Consider $$\sum_{n=2}^{\infty}\frac1{n\sqrt{\ln n}}.$$

1. Verify the domain and all Integral Test hypotheses for $f(x)=1/[x\sqrt{\ln x}]$.

2. Evaluate the improper integral using $u=\ln x$.

3. Determine convergence or divergence.

4. Place this series within the family $\sum 1/[n(\ln n)^p]$ and state the threshold value of $p$.

*Question 10 – Solution

Let $f(x)=1/[x\sqrt{\ln x}]$ for $x\ge2$. It is continuous and positive. Its denominator $x\sqrt{\ln x}$ is increasing, so $f$ is decreasing. Equivalently, $$f'(x)=-\frac{2\ln x+1}{2x^2(\ln x)^{3/2}}<0.$$

Put $u=\ln x$, so $du=dx/x$: $$\int_2^b\frac{dx}{x\sqrt{\ln x}} =\int_{\ln2}^{\ln b}u^{-1/2}\,du =2\sqrt{\ln b}-2\sqrt{\ln2}.$$ This tends to infinity, so the series diverges by the Integral Test.

More generally, $$\sum\frac1{n(\ln n)^p}$$ converges for $p>1$ and diverges for $p\le1$. Here $p=1/2$, which lies on the divergent side.