Comparison and Limit Comparison Tests — Question 10

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

1. Show why limit comparison with $1/n$ and $1/n^p$ does not immediately decide the series.

2. Select a natural logarithmic benchmark or use the Integral Test.

3. Classify the series and state an explicit tail bound.

*Question 10 – Solution

Relative to $1/n$, $$\frac{1/[n(\ln n)^2]}{1/n}=\frac1{(\ln n)^2}\to0,$$ which does not transfer the harmonic series’ divergence. For any $p>1$, the ratio to $1/n^p$ is $n^{p-1}/(\ln n)^2\to\infty$, which does not transfer that $p$-series’ convergence.

Let $f(x)=1/[x(\ln x)^2]$. It is positive, continuous, and decreasing for $x\ge2$. With $u=\ln x$, $$\int_2^{\infty}\frac{dx}{x(\ln x)^2} =\int_{\ln2}^{\infty}u^{-2}\,du=\frac1{\ln2}<\infty.$$ Thus the series converges by the Integral Test.

$$\boxed{\frac1{\ln(N+1)}\le R_N\le\frac1{\ln N}}.$$ The correct benchmark retains the logarithm rather than discarding it.