Convergence and Divergence of Series — Question 2

*Question 2 Determine whether $\displaystyle\sum_{n=2}^{\infty}\frac1{n\ln n}$ converges or diverges.

1. Verify that the terms approach zero and explain why that does not settle convergence.

2. Group the indices into dyadic blocks $2^k\le n<2^{k+1}$.

3. Find a lower bound for each block and use it to classify the series.

*Question 2 – Solution

For $n\ge2$, $a_n=1/(n\ln n)>0$, and $n\ln n\to\infty$. Thus $a_n\to0$. The nth-term test is inconclusive, so a tail estimate is required.

Use the block $2^k\le n<2^{k+1}$. It contains exactly $2^k$ integers. Within this block, $$n<2^{k+1},\qquad \ln n<(k+1)\ln2,$$ and hence $$\frac1{n\ln n}>\frac1{2^{k+1}(k+1)\ln2}.$$

Because the block contains $2^k$ terms, its sum is greater than $$2^k\frac1{2^{k+1}(k+1)\ln2}=\frac1{2(k+1)\ln2}.$$

Summing these lower bounds over $k$ produces the constant multiple $(2\ln2)^{-1}\sum 1/(k+1)$ of the divergent harmonic series. Hence the original partial sums are unbounded and $$\boxed{\displaystyle\sum_{n=2}^{\infty}\frac1{n\ln n}\text{ diverges}.}$$

The hidden continuous substitution is $u=\ln x$, since $du=dx/x$ and $\int dx/(x\ln x)=\ln(\ln x)+C$. This explains the exceptionally slow divergence; a large numerical cutoff can misleadingly suggest stabilization.