Integral Test — Question 3

*Question 3 Consider $$\sum_{n=2}^{\infty}\frac1{n(\ln n)^2}.$$

1. Verify that $f(x)=1/[x(\ln x)^2]$ is positive, continuous, and decreasing for $x\ge2$.

2. Apply the Integral Test, showing the substitution and evaluation of the improper integral.

3. Derive two-sided Integral Test bounds for the remainder $R_N$.

4. Briefly compare the result with $\sum 1/(n\ln n)$.

*Question 3 – Solution

Define $f(x)=1/[x(\ln x)^2]$ for $x\ge2$. It is continuous and positive, and $$f'(x)=-\frac{\ln x+2}{x^2(\ln x)^3}<0,$$ so it decreases.

Let $u=\ln x$, $du=dx/x$: $$\int_2^{\infty}\frac{dx}{x(\ln x)^2} =\int_{\ln2}^{\infty}u^{-2}\,du=\frac1{\ln2}<\infty.$$ Hence the series converges. The extra logarithm power changes $\int du/u$ into the convergent $p$-integral $\int du/u^2$.

$$\boxed{\frac1{\ln(N+1)}\le R_N\le\frac1{\ln N}}.$$ These bounds follow by integrating from $N+1$ and $N$, respectively.