Test $\sum_{n=2}^{\infty}{(-1)^n \over n(\ln{n})^p}$ for conditional and absolute convergence I test
$$\sum_{n=2}^{\infty}{(-1)^n \over n(\ln{n})^p}$$
for conditional and and absolute convergence.
I am able to conclude about the convergence of $\sum_{n=2}^{\infty}{1 \over n(\ln{n})^p}$ from the Cauchy condensation test. The series converges for $p>1$, so the original series converges absolutely for such $p$. 
Now, I would like to check conditional convergence. Let's consider $a_n={1 \over n(\ln{n})^p}$ again. For $p\ge 0$ we have $a_n$ decreasing to $0$ and from the Leibniz criterion we get conditional convergence for $p>1$. My problem is, how do we approach conditional convergence for $p<0$? My sense is that we have to show the terms of the series are larger than the terms of any well-known divergent series for sufficiently large $n$, however, I can't find such a series.
 A: Let $p<0$ and then $s=-p$ so that $s>0$.  Now we have
$$\sum_{n=1}^\infty\frac{(-1)^n(\ln n)^s}n$$
We know that $\sum_{n=2}^N(-1)^n$ is bounded and that $\lim_{n\to\infty}\frac{(\ln n)^s}n=0$ (you may evaluate with repeated L'Hospital's rule) and monotone for all $n=\lceil e^s\rceil$.
Thus, by the Dirichlet test, it converges.  Indeed, the alternating test will show it converges for any $s,p$.
It should also be clear that it is not absolutely convergent, as you may use comparison test to the harmonic numbers.
As a last note, your sum can be given by
$$\sum_{n=2}^\infty\frac{(-1)^n(\ln n)^s}n=-e^{-s\pi i}\eta^{(s)}(1)$$
which are fractional derivatives of the Dirichlet eta function evaluated at $1$.  If $s\in\mathbb N$, this can more simply be written as
$$\sum_{n=2}^\infty\frac{(-1)^n(\ln n)^s}n=(-1)^{s+1}\eta^{(s)}(1)$$
A: This is an answer 
to your comment
about showing that
${\ln{n} \over n} \to 0
$.
This is equivalent to
$n^{1/n} \to 1$.
This can be done with
Bernoulli's inequality
as follows:
$(1+\frac1{\sqrt{n}})^n
\ge 1+\frac{n}{\sqrt{n}}
> \sqrt{n}
= n^{1/2}
$.
Raising to the
$2/n$ power,
we get
$\begin{array}\\
(1+\frac1{\sqrt{n}})^2
&\gt n^{1/n}\\
\text{or}\\
n^{1/n}
&\lt 1+2\frac1{\sqrt{n}}+\frac1{n}\\
&\le 1+3\frac1{\sqrt{n}}\\
&\to 1\\
&\text{as }
n \to \infty\\
\end{array}
$
