Absolute Convergence and Divergence — Question 10

Question 10 – Solution

Step 1: Dominate the logarithm.

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

Step 2: Test absolute convergence.

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

Step 3: Classify.

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