Absolute and Conditional Convergence of Series Involving Double Log Test the following series for absolute and conditional convergence:
$$\sum_{k=3}^\infty \frac {(-1)^k \log k}{k\log (\log k)}. \tag{1} $$
What test do I have to use to test for convergence of this series?
Use of the double log always confuses me.
I have two more series:
$$\sum_{k=2}^\infty \frac{(-1)^k}{k^\alpha+(-1)^k}, \alpha \gt 0 \tag{2} $$
$$\sum_{k=1}^\infty (-1)^kk^{-\alpha}, \quad \alpha \in \mathbb{R} \tag{3}$$

$(3)$ is easy.
Convergence:
By alternating series, so $\lim_{k\rightarrow \infty} k^{-\alpha}$ if $\alpha \gt 0 $.
In this case, this series converges conditionally.
However the series does not converge absolutely unless $\alpha \gt 1$ by p-series.
Divergence:
If $\alpha \le 0$ this series does not converge absolutely or conditionally.
This is because $\lim_{k\rightarrow \infty} k^{-\alpha}$ is not equal to zero.

(Reference: Elementary Classical Anaylsis, Marsden, pg. 323)
 A: We have
$$\int_3^{L}\frac{\log x}{x\log (\log x)}dx=\int_{\log (\log 3)}^{\log (\log L)}\frac{e^{2u}}{u}du$$
which clearly diverges as $L \to \infty$.  Thus there series does not absolutely converge by the integral test.  Inasmuch as $\lim_{k\to \infty}\frac{\log k}{k\log(\log k)}=0$, the series conditionally coverges.

For the second  series 
$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k^{\alpha}+(-1)^k} \tag 1$$
we have
$$\lim_{k\to \infty }\frac{1}{k^{\alpha}+(-1)^k}= 0$$
for $\alpha >0$ only.  Thus, the series $(1)$ converges for $\alpha >0$.  Now, to test for absolute convergence, we have 
$$\left|\frac{(-1)^k}{k^{\alpha}+(-1)^k}\right|<\frac{1}{k^{\alpha}-1}$$
and thus, the series converges absolutely for $\alpha >1$.
NOTE:  The third series is effectively the same as the second one.
A: The series is conditionally convergent by Dirichlet's test, since the partial sums of $(-1)^k$ are bounded and $\frac{\log k}{k \log(\log k)}$ is eventually decreasing to zero. It is not absolutely convergent by comparison with the harmonic series, since for any $k$ big enough:
$$\frac{\log k}{k\,\log\log k} > \frac{\log k}{k\,\log k} = \frac{1}{k}.$$
You may use similar arguments for dealing with the series in the second part, too.
