Strategy for Series — Question 2

*Question 2 Determine whether $\displaystyle\sum_{n=2}^{\infty}\frac1{n\log n}$ converges or diverges. Explain why the Integral Test is especially well suited to this expression.

*Question 2 – Solution

Let $f(x)=1/(x\log x)$ for $x\ge2$. It is positive and continuous. Also, $$f'(x)=-\frac{\log x+1}{x^2(\log x)^2}<0,$$ so $f$ is decreasing.

The integrand contains the differential $dx/x=d(\log x)$, suggesting $u=\log x$: $$\int_2^b\frac{dx}{x\log x}=\int_{\log2}^{\log b}\frac{du}{u}=\log(\log b)-\log(\log2).$$ As $b\to\infty$, this tends to $\infty$.

The series diverges by the Integral Test. The substitution exposes the antiderivative $\log(\log x)$ immediately, which is why this test is more natural than a ratio or root test here.