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 Post subject: On a series with cosinePosted: Sat May 20, 2017 6:18 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Evaluate the series

$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \cos \left( \frac{n \pi}{12} \right)}{n 3^n}$$

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 Post subject: Re: On a series with cosinePosted: Mon May 22, 2017 4:01 pm

Joined: Tue May 10, 2016 3:56 pm
Posts: 33
Recall that
$$\sum_{n=1}^\infty \frac{1}{n}\,x^n = - \ln(1-x).$$
Let $i = \sqrt{-1}$. Then
\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{n 3^n}\,\cos\left(\frac{n\pi}{12}\right) & = & \Re\left\{\sum_{n=1}^\infty \frac{1}{n}\,\left(\frac{e^{i\pi/12}}{3}\right)^n\right\}\\
& = & \Re\left\{-\ln\left(1 - \frac{e^{i\pi/12}}{3}\right)\right\}\\
& = & \ln\frac{3}{\sqrt{10 - 3\sqrt{2 + \sqrt{3}}}}\\
& = & \ln 3 - \frac{1}{2}\,\ln(10 - 3\sqrt{2 + \sqrt{3}}).
\end{eqnarray*}

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 Post subject: Re: On a series with cosinePosted: Tue May 23, 2017 2:28 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
Or by recalling that:

$$\sum_{n=1}^{\infty} \frac{x^n \cos na}{n} = -\frac{1}{2} \log \left( x^2 - 2x \cos a + 1 \right) \; , \; |x|<1$$

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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