Comparison and Limit Comparison Tests — Question 9

*Question 9 Consider $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^3}$.

1. Prove a usable power bound for $\ln n$.

2. Compare the summand with a convergent $p$-series.

3. Explain why limit comparison with $1/n^3$ is inconclusive.

*Question 9 – Solution

We claim $\ln x\le\sqrt x$ for $x\ge1$. Let $h(x)=\sqrt x-\ln x$. Then $$h'(x)=\frac{\sqrt x-2}{2x},$$ so the minimum occurs at $x=4$, where $h(4)=2-\ln4>0$. The claim follows.

For $n\ge1$, $$0\le\frac{\ln n}{n^3}\le\frac{\sqrt n}{n^3}=\frac1{n^{5/2}}.$$ The benchmark is a convergent $p$-series with $p=5/2>1$, so Direct Comparison proves convergence.

Comparison with $1/n^3$ by ratios gives $(\ln n/n^3)/(1/n^3)=\ln n\to\infty$, outside the standard finite-positive Limit Comparison case. A slightly larger convergent power benchmark is more useful.