Integral Test — Question 2

*Question 2 Consider the logarithmic series $$\sum_{n=2}^{\infty}\frac1{n\ln n}.$$

1. Explain why the series begins at $n=2$, and verify all three Integral Test hypotheses for $f(x)=1/(x\ln x)$.

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

3. State the convergence or divergence conclusion and describe the growth rate of the truncated integral.

The nth-term test alone is not a complete solution.

*Question 2 – Solution

Let $f(x)=1/(x\ln x)$ for $x\ge2$. It is continuous and positive because $x>1$. Moreover, $$f'(x)=-\frac{\ln x+1}{x^2(\ln x)^2}<0,$$ so $f$ is decreasing, and $f(n)$ equals the summand.

With $u=\ln x$ and $du=dx/x$, $$\int_2^b\frac{dx}{x\ln x}=\int_{\ln2}^{\ln b}\frac{du}{u} =\ln(\ln b)-\ln(\ln2).$$ As $b\to\infty$, $\ln(\ln b)\to\infty$. Thus the improper integral diverges.

All hypotheses hold, so the Integral Test gives $$\boxed{\sum_{n=2}^{\infty}\frac1{n\ln n}\text{ diverges}.}$$ The growth scale $\ln\ln b$ explains why divergence is extremely slow.