Comparison and Limit Comparison Tests — Question 10

Question 10 – Solution

Step 1: Audit ordinary power benchmarks.

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.

Step 2: Use the natural logarithmic test.

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.

Step 3: State the tail estimate.

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