How do you solve this series question? : $\cos (n\pi )/ \ln(6n) $ The problem is, Select the FIRST correct reason on the list why the given series converges.
A. Geometric series 
B. Comparison with a convergent p series
C. Integral test 
D. Ratio test 
E. Alternating series test 
I understand all the other sub problems except for this one:
$$\sum_1^\infty \frac{\cos(n\pi)}{\ln(6n)}$$ 
The answer is E, Alternating series test. The alternating series test is what you do if an alternating series doesn't converge absolutely using the absolute convergence test right? Anyways, I'm a little stumped on how to solve this problem. 
My first thought was to take the absolute value of it, which means $\cos(n\pi)\leq 0$. Therefore the original sum was $\leq \frac{1}{\ln(6n)}$ ... Bla! 
Any help would be very much appreciated!
 A: Edit in response to Robert Israel's comment.
Since $\cos (n\pi )=(-1)^{n}$, the series is
$$\sum_{n=1}^{\infty }\frac{(-1)^{n}}{\ln (6n)},$$
which converges by the alternating series test because $\frac{1}{\ln (6n)}\rightarrow 0$ and $1/\ln(6n)$ is monotone decreasing.
But
$$\sum_{n=1}^{\infty }\frac{1}{\ln (6n)}$$
is not convergent because: 


*

*so is
$$\int_{1}^{\infty }\frac{dx}{\log (6x)},$$
as can be seen by the limit test with $$\int_{1}^{\infty }\frac{dx}{6x},$$
applying L'Hôpital's rule. 

*Or by a direct comparison test with the divergent series $\sum_{n=1}^{\infty }\frac{1}{6n}$ 
$$\sum_{n=1}^{\infty }\frac{1}{\ln (6n)}\ge\sum_{n=1}^{\infty }\frac{1}{6n},$$ 
because the harmonic series $\sum_{n=1}^{\infty }\frac{1}{n}$ is divergent.
A: The expected answer is of course E.  But for fun we show that things are slightly more complicated than that.  If the question were "what is the simplest test to use,"
the answer would be unambiguously E. But the question is about the first test in the list. A somewhat tongue in cheek case for C is made below.  
Our series is
$$\sum_{n=1}^\infty \frac{(-1)^n}{\ln(6n)}.$$ 
For convenience, change all the signs.  So we are interested in
$$\left(\frac{1}{\ln(6)}-\frac{1}{\ln(12)}\right)+ \left(\frac{1}{\ln(18)}-\frac{1}{\ln(24)}\right)+ \left(\frac{1}{\ln(30)}-\frac{1}{\ln(36)}\right)+ \cdots.$$
Each group has shape
$$\frac{1}{\ln(12n+6)}-\frac{1}{\ln(12n+12)}.$$
Bring to a common denominator, then simplify the top. We get
$$\frac{\ln\left(1+\frac{1}{2n+1}\right)}{\ln(12n+6)\ln(12n+12)}.$$
But if $x$ is small positive, we have $0<\ln(1+x)<x$.
Now an easy integral test comparison with
$$\int_2^\infty \frac{dx}{x(\ln x)^2}$$
shows that we have convergence.  
