Does $\sum_{n=2}^\infty \frac{1}{n\log(n)}$ converge or diverge? 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? 
 A: Diverges by Integral test, since $$\displaystyle \int_2^\infty \dfrac{dt}{t \ln(t)} = \int_2^\infty \dfrac{d}{dt} \ln(\ln(t))$$ diverges.
A: 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.
A: 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
$\ln(1+x) < x$.
