Convergence of $\sum_{n=3}^\infty \frac {1}{n \ln n}$ I know I have seen something similar and there is a telescoping trick to the convergence but it is eluding me.


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.

