Does $\sum_{n=2}^\infty \frac{1}{n\log(n)}$ converge or diverge?

Question

So I know that $\sum_{n\in\mathbb{N}}1/n$ diverges and $\sum_{n\in\mathbb{N}}1/n^2$ converges. What about the series $\sum_{n=2}^\infty1/n(\log(n))$? I'm pretty confident that it diverges but is there a quick justification?

Answer 1

Diverges by Integral test, since $$\displaystyle \int_2^\infty \dfrac{dt}{t \ln(t)} = \int_2^\infty \dfrac{d}{dt} \ln(\ln(t))$$ diverges.

Answer 2

HINT

By Cauchy condensation test we can consider the condensed series

$$\sum_{n=2}^\infty \frac{1}{n\log(n)} \rightarrow\sum_{n=2}^\infty \frac{2^n}{2^n\log(2^n)}$$

and the first converges if and only if the second one converges.

By the same test we can also determine the convergence for the more general case

$$\sum_{n=2}^\infty \frac{1}{n^a\log^b(n)}$$

known as Bertrand Series.

Answer 3

I'll see if I can come up with a discrete version of the integral test.

$\begin{array}\\ \ln(\ln(t+1))-\ln(\ln(t)) &=\ln\left(\dfrac{\ln(t+1)}{\ln(t))}\right)\\ &=\ln\left(\dfrac{\ln(t)+(\ln(t+1)-\ln(t))}{\ln(t))}\right)\\ &=\ln\left(1+\dfrac{\ln(t+1)-\ln(t)}{\ln(t))}\right)\\ &=\ln\left(1+\dfrac{\ln(\frac{t+1}{t})}{\ln(t))}\right)\\ &=\ln\left(1+\dfrac{\ln(1+\frac{1}{t})}{\ln(t))}\right)\\ &<\ln\left(1+\dfrac{\frac{1}{t}}{\ln(t))}\right) \qquad\text{since } \ln(1+x) < x \text{ for } 0 < x\\ &=\ln\left(1+\dfrac{1}{t\ln(t))}\right)\\ &<\dfrac{1}{t\ln(t))}\\ \end{array} $

so

$\begin{array}\\ \sum_{k=2}^n \dfrac{1}{k\ln(k))} &>\sum_{k=2}^n (\ln(\ln(k+1))-\ln(\ln(k)))\\ &=\ln(\ln(n+1))-\ln(\ln(2)))\\ &\to \infty\\ \end{array} $

Yep - that works.

By iterating this, we can show that $\sum_n \dfrac1{n\ln(n)\ln(\ln(n))\ln(\ln(\ln(n)))...} $ diverges.

All that is needed is $\ln(1+x) < x$.