Determining if a series converges conditionally or absolutely Determine whether the following series converges conditionally, or converges absolutely. $$\sum^\infty_{k=2}\frac{\sin(\frac{\pi}{2}+k\pi)}{\sqrt{k}\ln(k)}$$
What could I use here to work this out? This isn't monotone. I've tried using the Ratio Test but this seems kinda cumbersome. Is there a better way to do this?
Was I even allowed to use the Ratio Test? I've just remembered it's not for $\sum^\infty_{n=c}$ but for $n=1$. 
 A: Note that
$$\sum^\infty_{k=2}\frac{\sin(\frac{\pi}{2}+k\pi)}{\sqrt{k}\ln(k)}=\sum^\infty_{k=2}\frac{(-1)^k}{\sqrt{k}\ln(k)}$$
thus it converges conditionally by alternating series test  while
$$\sum^\infty_{k=2}\left|\frac{\sin(\frac{\pi}{2}+k\pi)}{\sqrt{k}\ln(k)}\right|=\sum^\infty_{k=2}\frac{1}{\sqrt{k}\ln(k)}$$
diverges by limit comparison test with $\frac{1}{k^\frac34}$.
For the latter, as an alternative suggested by Mark Viola, note that since for any $a>0$
$$\log x^a\le x^a-1 \implies\log x\le \frac{x^a-1}{a}<\frac{x^a}{a}$$
selecting $a=\frac12$ we have
$$\sum^\infty_{k=2}\frac{1}{\sqrt{k}\ln(k)}>\sum^\infty_{k=2}\frac{1}{2k}$$
A: Note that $\sin(\pi/2 +k\pi)$ just alternates between $1$ and $-1,$ so has absolute value $1$. So absolute convergence is just about $\frac{1}{\sqrt k \ln k}.$ For this I recommend a comparison with $1/k$. For the convergence of the series itself I recommend the alternating series test.
A: Firstly, we will study if the serie
$$\sum_{k=2}^\infty\left|\frac{\sin\left(\frac{\pi}{2}+k\pi\right)}{\sqrt{k}\ln(k)}\right|$$
converges, because if it does, directly we have that our original one does too. 
Note that 
$$\sin\left(\frac{\pi}{2}+k\pi\right)=(-1)^{k}$$
so we have that 
$$\sum_{k=2}^\infty\left|\frac{\sin\left(\frac{\pi}{2}+k\pi\right)}{\sqrt{k}\ln(k)}\right|=\sum_{k=2}^\infty\left|\frac{1}{\sqrt{k}\ln(k)}\right|=\sum_{k=2}^\infty\frac{1}{\sqrt{k}\ln(k)}$$
Now, we're going to use the integral test for convergence
https://en.wikipedia.org/wiki/Integral_test_for_convergence
As the integral 
$$\int \frac{1}{\sqrt{x}\ln(x)}dx=\int_{\ln(2)}^\infty  \frac{e^t}{t\sqrt{e^t}}dt =\int_{\ln(2)}^\infty  \frac{1}{t\sqrt{e^{-t}}}dt $$
has not a primitive on elementary function, we can proceed comparing this integral with another and apply the comparison test for improper integrals 
http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx
Comparing it  with $g(t)=\frac{1}{t}$, we get that
$$\lim_{t\to \infty} \frac{\frac{1}{t\sqrt{e^{-t}}}}{\frac{1}{t}}=\lim_{t\to \infty} \frac{1}{\sqrt{e^{-t}}}=+\infty$$
and as $g$ diverges, $f$ does it too. 
