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!
Best Answer
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.