$\sum_{j=3}^\infty \frac{1}{j(\log(j))^3}$ converges or diverges?

Question

Showing that

$\sum_{j=3}^\infty \frac{1}{j(\log j)^3}$

diverges or converges

How would it converges or diverges. I thought about using comparison test buy $\frac{1}{j}$ is bigger than the sum.

What is the key the log seems to get my stuck.

Answer 1

Using the Cauchy condensation test, the convergence/divergence properties of your series are the same as those of

$$\sum_{n=1}^\infty \frac{2^n}{2^n \log(2^n)^3} = C \sum_{n=1}^\infty \frac{1}{n^3}.$$

A related trick is "integration by Riemann sums at dyadic points", which can be used in dealing with convergence/divergence of similar integrals.

Answer 2

By using the integral test, your series and the following integral are of the same nature: $$ 0<\int_3^{\infty}\frac1{x (\log x)^3}dx=\int_3^{\infty}(\log x)^{-3}(\log x)'dx=\left[- \frac1{2 (\log x)^2}\right]_3^\infty=\frac1{2 (\log 3)^2}<\infty. $$ Your series is thus convergent.