Does the series $\sum_{n = 2}^\infty \frac{1}{n^a\ln n}$, $a > 1$ converge or diverge? Here are the approaches I tried.


*

*If I use the limit comparison test with $\frac{1}{n^a}$, I get $\lim\limits_{n \to \infty}\frac{1}{\ln n} = 0$, which I don't think is very helpful.

*I can't use the direct comparison test with $\frac{1}{n^a}$ either because the given series is greater than it. 

*I don't think I can use the integral test either. I don't know how to integrate $\int_0^{\infty}\frac{1}{n^a\ln n} dn$.

*I think I might be able to compare the series with $\frac{1}{n^{a + 1}}$, since $\ln n < n$. 

 A: For $n\geq 3$ we have that
$$
n^a\ln n\geq n^a
$$
whence
$$
0\leq \frac{1}{n^a\ln a}\leq\frac{1}{n^a}
$$
and you can conclude using comparison to a p-series.
A: Try Cauchy condensation test.
Here is a link.
The sum converges if $\sum 2^n\cdot a_{2^n}$ converges.
$\sum_{n=0}^{\infty} 2^n\cdot \frac{1}{(2^n)^aln(2^n)}=\frac{1}{2^aln(2)}\sum_{n=0}^{\infty} \frac{1}{n\cdot2^n}$
It is easy to show that $\sum_{n=0}^{\infty} \frac{1}{n\cdot2^n}$ converge.
A: The following series is a kind of $p$ - series:$$\sum_{n = 2}^\infty \frac{1}{n^a\ln^p n}$$ which converges if \begin{cases} a>1, \ p\in \mathbb R\\ a=1, \ p>1\end{cases}
and diverges otherwise.
Thus your series converges for $a>1$. Also, like Foobaz John said:
$$0\leq \frac{1}{n^a\ln a}\leq\frac{1}{n^a}$$ 
and
$$\sum_{n = 2}^\infty \frac{1}{n^a} \text{ converges for } a>1$$
A: As this is a series with positive terms, a simple comparison test will do: for all $n>2$, 
$$0<\frac1{n^a\log n}<\frac 1{n^a}$$
and the latter is a convergent Riemann series.
Note: the same test shows that $\;\displaystyle\sum_{n\ge 2}\frac1{n^a\log^bn}\;$ converges for any $b\in\mathbf R$ if $a>1$.
