Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$
Check the conditions fit for the Condensation Test :
$$a_n:=\frac1{n\log n}\implies 2^na_{2^n}=\frac{2^n}{2^n\log2^n}=\frac1{\log 2}\frac1n$$
and since the series of the rightmost sequence is just a multiple of the harmonic series and thus diverges, also our series diverges.
To see whether $\sum_2^\infty 1/(n \log n)$ converges, we can use the integral test. This series converges if and only if this integral does: $$ \int_2^\infty \frac{1}{x \log x} dx = \left[\log(\log x)\right]_2^\infty $$ and in fact the integral diverges.
This is part of a family of examples worth remembering. Note that $$ d/dx \log(\log(\log x)) = d/dx \log(\log x) \cdot \frac{1}{\log (\log x)} = \frac{1}{x \log x \log(\log x)} $$ and $\log (\log (\log x)) \to \infty$ as $x \to \infty$ hence $\sum \frac{1}{n \log n \log (\log n)}$ diverges as well. Similarly, by induction we can put as many iterated $\log$s in the denominator as we want (i.e. $\sum \frac{1}{n \log n \log(\log n) \ldots \log (\ldots (\log n) \ldots )}$ where the $i$th log is iterated $i$ times), and it will still diverge. However, as you should check, $\sum \frac{1}{x \log^2x}$ converges, and in fact (again by induction) if you square any of the iterated logs in $\sum \frac{1}{n \log n \log(\log n) \ldots \log (\ldots (\log n) \ldots )}$ the sum will converge.
We circumvent using the integral test or its companion, the Cauchy condensation test. Rather, we use creative telescoping to show that the series $\sum_{n=3}^\infty \frac{1}{n\log(n)}$ diverges. To that end, we now proceed.
We will use the well-known inequalities for the logarithm (SEE THIS ANSWER)
$$\frac{x-1}{x} \le \log(x)\le x-1 \tag1$$
Using the right-hand side inequality in $(1)$, we see that
$$\log\left(\frac{n+1}{n}\right)\le \frac1n \tag 2$$
and
$$\log\left(\frac{\log(n+1)}{\log(n)}\right)\le \frac{\log(n+1)}{\log(n)}-1 \tag3$$
Applying $(2)$ and $(3)$ yields
$$\begin{align} \sum_{n=3}^N \frac{1}{n\log(n)} &\ge \sum_{n=3}^N \frac{\log\left(\frac{n+1}{n}\right)}{\log(n)}\\\\ &=\sum_{n=3}^N \left(\frac{\log(n+1)}{\log(n)} -1\right)\\\\ &\ge \sum_{n=3}^N \log\left(\frac{\log(n+1)}{\log(n)}\right)\\\\ &=\sum_{n=3}^N \left(\log(\log(n+1)) -\log(\log(n)) \right)\\\\ &=\log(\log(N+1))-\log(\log(3)) \end{align}$$
Inasmuch as $\lim_{N\to \infty}\log(\log(N+1))=\infty$, the series of interest diverges by comparison.
And we are done!
TOOLS USED: The right-hand side inequality in $(1)$ and summing a telescoping series.