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.
 A: 
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.

A: For a decreasing function $a(x) \ge 0$, there is a theorem known as the Integral Test, which says that $\int_{m}^\infty a(x)$ converges if, and only if, $\sum_{n=m}^\infty a(n)$ converges. This is the (probably best) way of determining convergence or divergence here - you'll want to take $a(x) = \frac{1}{x\ln(x)}$.
A: The easiest way to do this is use Cauchy...$f(n)$ coverges or diverges so as $2^nf(2^n)$ converges or diverges.But then the latter is $\Sigma$ $1/nln(2)$ which diverges.  
