On a series with cosine

Calculus (Integrals, Series)
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

On a series with cosine

#1

Post by Tolaso J Kos »

Evaluate the series

$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \cos \left( \frac{n \pi}{12} \right)}{n 3^n}$$
Imagination is much more important than knowledge.
mathofusva
Posts: 33
Joined: Tue May 10, 2016 3:56 pm

Re: On a series with cosine

#2

Post by mathofusva »

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*}
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: On a series with cosine

#3

Post by Riemann »

Or by recalling that:

$$\sum_{n=1}^{\infty} \frac{x^n \cos na}{n} = -\frac{1}{2} \log \left( x^2 - 2x \cos a + 1 \right) \; , \; |x|<1 $$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 8 guests