Absolute Convergence and Divergence — Question 10

*Question 10 Consider $\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n^2}$.

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

2. Use it to test absolute convergence.

3. Classify the original series and explain why the AST is not needed.

*Question 10 – Solution

For $x\ge1$, $\ln x\le\sqrt x$. One proof minimizes $h(x)=\sqrt x-\ln x$; its minimum is $h(4)=2-\ln4>0$.

$$0\le\frac{\ln n}{n^2}\le\frac{\sqrt n}{n^2}=\frac1{n^{3/2}}.$$ The upper benchmark is a convergent $p$-series, so Direct Comparison proves convergence of the absolute-value series.

The original series converges absolutely. The AST could prove ordinary convergence after checking eventual decrease, but absolute comparison is faster and stronger.