convergence of $n \sum_{k=n}^\infty \frac{1}{k^2 \log(k)}$

Question

I am stuck in a calculus step of an exercise from Durett's book.

I am trying to show that $n \sum_{k=n}^\infty \frac{1}{k^2 \log(k)}$ goes to zero when $n$ goes to infinity. I understand that the sum is convergent hence the rest of the series goes to zero but I can't prove that it goes to zero quicker than n goes to infinity .. Since I don't know any primitive of $\frac{1}{k^2 \log(k)}$, my attempts with comparison series/integral have been unsuccessful. Maybe I missed a calculus theorem.

Thank you in advance for your hints.

Answer 1

You can use the integral test to show that

$$\sum_{k=n}^\infty{1\over k^2}={1\over n^2}+\sum_{k=n+1}^\infty{1\over k^2}\lt{1\over n^2}+\int_n^\infty{dx\over x^2}\lt{1\over n^2}+{1\over n}\le{2\over n}$$

It follows that

$$n\sum_{k=n}^\infty{1\over k^2\log k}\lt{n\over\log n}\sum_{k=n}^\infty{1\over k^2}\lt{2\over\log n}\to0$$

Answer 2

Actually it is enough to apply the idea behind Cauchy's condensation test: $$ \frac{2^{m}}{2^{2m+2}\log(2^{m+1})}\leq\sum_{2^m\leq k < 2^{m+1}}\frac{1}{k^2 \log k}\leq \frac{2^m}{2^{2m}\log(2^m)} $$ to get $$ \sum_{2^m\leq k < 2^{m+1}}\frac{1}{k^2\log k}=\Theta\left(\frac{1}{m 2^m}\right) $$ and $$ \sum_{k\geq x}\frac{1}{k^2\log k}=\Theta\left(\frac{1}{x\log x}\right).$$