# 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!

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The value of $\cos(n\pi)$ is actually rather simple, you might start by computing it. –  Did Jul 22 '11 at 23:07
crap! didn't see that. But that leaves it to (-1)^n / ln(6n) still right? the absolute value is 1/ln(6n)... which diverges, but i don't know how to prove it –  user13327 Jul 22 '11 at 23:13
I figured it out! 1/ln(6x) ~ 1 / ln(x) > 1 / n which diverges by the p test. Yay! –  user13327 Jul 22 '11 at 23:37
Since $\cos (n\pi )=(-1)^{n}$, the series is $\sum_{n=1}^{\infty }\frac{(-1)^{n}}{\ln (6n)}$ –  Américo Tavares Jul 23 '11 at 0:05
The divergence of the series $\Sigma 1/\ln(6n)$ is somewhat moot here; it's true that that series diverges (and you can convince yourself that A and B don't apply easily, and that D doesn't apply with a little bit more effort. You do need the divergence of that series to show that the interval test doesn't work, but the reason why the alternating series test works is simply that the sequence $1/\ln(6n)$ converges monotonically to zero. –  Steven Stadnicki Jul 23 '11 at 0:05

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:

1. 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.

2. 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.

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There's more to the alternating series test than $\frac{1}{\ln(6n)} \to 0$: you also need the fact that this sequence is monotonic. And you don't need to worry about integration: you can compare $\sum_{n=1}^\infty \frac{1}{\ln(6n)}$ to $\sum_{n=1}^\infty \frac{1}{6n}$ directly, if you know that the harmonic series diverges. –  Robert Israel Jul 23 '11 at 0:32
@Robert Israel: Thanks! I corrected the test for the 1st series and added the simpler test for the 2nd. –  Américo Tavares Jul 23 '11 at 0:51
Thank you, all of you! –  user13327 Jul 23 '11 at 3:39

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.

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