When does $\sum\frac{1}{(n\ln n)^a}$ converge? 
I've to investigate when does $\displaystyle \sum\frac{1}{(n\ln n)^a}$ converge?

For $a>1$, I used the limit comparison test as $\displaystyle \frac{\frac{1}{(n\ln n)^a}}{\frac{1}{n^a}} \to 0$ as $n\to \infty$ so that $\displaystyle \sum\frac{1}{(n\ln n)^a}$ converges together with $\displaystyle\sum \frac{1}{n^a}$ and for $a\le\frac{1}{2}$ notice that $\displaystyle \sum\frac{1}{(n\ln n)^a}>\sum \frac{1}{n^{2a}}$ and since RHS diverges we can see $\displaystyle \sum\frac{1}{(n\ln n)^a}$ diverges . But what happens if $a\in(\frac{1}{2},1)$? Or is there any other method to cover these cases on one step?
 A: Here is an approach. By the Cauchy condensation test we can study instead the series 

$$ \sum_{n} \frac{2^{(1-a)n}}{n^a} = \sum_{n} \frac{z^n}{n^a}, $$

where $z=2^{(1-a)}$. The last series has the radius of convergence $|z|<1$ which implies 

$$ 2^{1-a} < 1 \implies 1-a<0 \implies a > 1 $$ 

Cauchy Condensation Theorem:

For a positive non-increasing sequence $f(n)$, the sum
  $$ \sum_{n=1}^{\infty}f(n)  $$
  converges if and only if the sum
  $$\sum_{n=0}^{\infty} 2^{n}f(2^{n})$$
  converges. 

A: Hint:  Use the integral test. 
A: You've already handled $a>1$, so look at $a \leq 1$. Deal with $a=1$ using the integral test; you should find that the series diverges. Consequently you can show the case where $a<1$ diverges by direct comparison to the case where $a=1$.
A: Lemma:
A series $\sum a_k$ whose terms are positive and monotonously decreasing is convergent iff the series $\sum_{k=1}^\infty 2^k a_{2^k}$ converges.
Sketch of the proof :
$a_i$ is decreasing so
$a_1 + a_2 +(a_4 + a_4 )+ (a_8 + a_8 + a_8 +a_8) + \dots$ $\leq a_1 + a_2 +(a_3 +a_4) + (a_5 + a_6 +a_7 + a_8) +..$
Again,
$a_1 + (a_2 +a_3) + (a_4 + a_5 + a_6 + a_7)+\dots \leq a_1 +(a_2+a_2)+ (a_4+a_4+a_4+a_4)+\dots $
Now applying this lemma, we need to find the condition for which $$\sum\frac{2^k}{{(2^kk \ln2)}^a}$$ which is quite easy to find.
