Integral test for convergence Show that 
$$\sum_{n=3}^\infty \frac1{n \log n(\log \log n)^p}$$
converges if and only if $ p > 1$.
I know the hint for this: we have to use integral test so we know that the sum of $f(n)$ converges $\iff$  integral of $f(x)$ converges.
Anyone know how exactly to do this?
 A: Note that
$$
\int\limits_3^{+\infty}\frac{dx}{x\log x(\log\log x)^p}
=\int\limits_3^{+\infty}\frac{d(\log x)}{\log x(\log\log x)^p}
=\int\limits_3^{+\infty}\frac{d(\log\log x)}{(\log\log x)^p}
=\frac{(\log\log x)^{-p-1}}{-p-1}\Biggl|_3^{+\infty}
=\lim\limits_{x\to+\infty}\frac{(\log\log x)^{-p-1}}{-p-1}-\frac{(\log\log 3)^{-p-1}}{-p-1}=+\infty
$$
Hence by integral test the series
$$
\sum\limits_{n=3}^{+\infty}\frac{1}{n\log n(\log\log n)^p}
$$
diverges
A: Using Cauchy condensation test,
For $p=1$,
Here $$a_n=\frac{1}{n\log n \log (\log n)}$$
By Cauchy condensation test $\sum a_n $& $\sum 2^n a_{2^n}$ converges or diverges together,
\begin{align*}
\sum 2^n a_{2^n}&=\sum 2^n \frac{1}{2^n \log {2^n} \log (\log 2^n)}\\ 
                &=\sum \frac{1}{n\log 2 \log (n \log 2)} 
\end{align*}
Since, 
$\log 2 < 1$
$n \log 2 < n$
$\displaystyle \frac {1}{n \log 2} > \frac {1}{n}$
$\displaystyle \frac{1}{\underbrace{n \log 2}_{m} \log\underbrace{(n \log 2)}_{m}} > \frac{1}{n \log n}$
Check the behavior  $\displaystyle \sum \frac{1}{m \log m} $ it is divergent for p=1,
There is one example which you have to solve,
$$\sum_{n=2}^\infty \frac{1}{n(\log n)^p} \ \  \text{if} \ \  p>0 $$
For $P>1$ converges and $P\leq 1$ diverges. 
$\displaystyle \therefore \sum 2^n a_{2^n}$ is divergent
$\displaystyle \therefore \sum_{n=3}^\infty \frac{1}{n \log n \log (\log n)} $ is divergent.
I hope you can conclude behavior for $P>1$.        
