$\sum_{n=2}^{\infty} \frac{1}{n \log n}$ Prove series diverge using comparsion test . prove $$\sum_{n=2}^{\infty} \frac{1}{n \log n}$$ diverge. I have done this problem using cauchy integral test and condensation test. But i want to do it by comparison test or by limit comparison test . any hint about that .
Thanks in advanced.
 A: Why you would like to do this by comparison test (comparing with a series)?  Note that the "condensation test" is in some sense also a comparison test.
One straightforward way to do this, is to use the integral comparison test: The function $$f(x):= \frac{1}{x\log(x)}$$ 
is for $x > 2$ decreasing. Thus we get
$$\sum_{n=2}^m \frac{1}{n \log(n)} \ge \int_{3}^{m} \frac{1}{x \log(x)} \, d x = \log \log m - \log \log (3)$$
The same argument can be used to find an upper bound. This gives the asymptotic formula
$$\sum_{n=2}^m \frac{1}{n \log(n)} \sim \log \log m.$$
A: Define $a_n=(1/(n\log n)$ and $b_n=a_{2^m}$ where $2^m$ is the smallest power of $2$ greater than or equal to $n$.  The number of $b_n$ terms having a given value of $m$ is then $2^{m-1}$ and the sum of those terms is
$$(2^{m-1})\dfrac{1}{2^m\log(2^m)}=\dfrac{1}{2m\log 2}.$$
Repeated over all values of $m\ge 1$ this gives a constant times the divergent harmonic series for the $b_n$ series.  Since the latter diverges and $a_n\ge b_n>0$ by construction, the $a_n$ series also diverges.
This form of comparison test is called the Cauchy Condensation Test.  Valid for any monotonically decreasing series of positive terms, the test can be used to establish either convergence or divergence.  It is best known for proving the divergence if the harmonic series itself.
