Series, is this reasonable?

**Twig** · Aug 10th 2009, 03:19 PM

Hi!

$\sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k}))$ , convergent or divergent?

I tried:

$a_{k}=ln(1+(cos(\frac{\pi}{k}) -1 )$

Now using the Taylor expansion of $ln(1+x) = x-\frac{x^2}{2}+...$ with $x=(cos(\frac{\pi}{k})-1)$ AND using the taylor expansion of $cos(x)=1-\frac{x^2}{2}+...$ I got

$a_{k}=\frac{-\pi^{2}}{2k^{2}}+O(\frac{1}{k^{4}})$

To be able to use the comparison critera, $\lim_{k\to \infty} \frac{a_{k}}{b_{k}} = A > 0$ , I applied it with $b_{k}=\frac{1}{k^{2}}$ on $- \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k}))$

And hence $a_{k}=\frac{\pi^{2}}{2k^{2}}-O(\frac{1}{k^{4}})$

And $\lim_{k\to \infty} \frac{a_{k}}{b_{k}} = \frac{\pi^2}{2}>0$

And $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges.

So the original series converges.

I assume if $\sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) =s$ then $- \sum_{k=3}^{\infty} ln(cos(\frac{\pi}{k})) =-s$ ?

Basically what I want, is someone to check the work.

Thx!

**matheagle** · Aug 10th 2009, 11:35 PM

I ignored your Taylor work, but I tested your $1/n^2$ idea

I just tried the limit comp test with $\sum{\pi^2\over n^2}<\infty$

${\ln\cos (\pi/n)\over{\pi^2\over n^2}}\to -.5$ by letting $w=\pi/n\to 0^+$

By that, I mean $\lim_{w\to 0^+} {\ln\cos w\over w^2} = \lim_{w\to 0^+} {-\tan w\over 2w} =-.5$

So, $-\sum \ln\cos (\pi/n)$ is a positive series and it can be compared to the p=2 series.

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