My Math Forum Product-Series Computation

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 July 16th, 2012, 03:35 PM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Product-Series Computation [color=#000000]Prove that $\sum_{n=1}^{\infty}\prod_{k=1}^{n}\cos\left(\frac{ k\pi}{n}\right)= -0.8$.[/color]
 July 17th, 2012, 12:41 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Product-Series approximation I tried to solve. but my answer is -1.31... would you mind pm me the method? I am really curios.
 July 17th, 2012, 06:14 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Product-Series Computation $\sum_{n=1}^{\infty}\prod_{k=1}^{n}\cos$$\frac{k\pi }{n}$$=\sum_{n=1}^{\infty}$$-\frac{\sin\(\frac{\pi}{2}n$$}{2^{n-1}}\)=4\sum_{n=1}^{\infty}$$-\frac{1}{4}$$^n=-\frac{4}{5}$ I was aided largely by the identity: $\prod_{k=1}^{n-1}\cos$$\frac{k\pi}{n}$$=\frac{\sin$$\frac{\pi}{2} n)}{2^{n-1}}$  July 17th, 2012, 06:24 PM #4 Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Product-Series Computation [color=#000000]Yes Mark, I will post a more detailed solution tomorrow. [/color]  July 17th, 2012, 06:26 PM #5 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Product-Series Computation I look forward to it!  July 19th, 2012, 12:48 AM #6 Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Product-Series Computation [color=#000000]$\sum_{n=1}^{\infty}\left(\prod_{k=1}^{n}\cos\left( \frac{k\pi}{n}\right)\right)=-\sum_{n=1}^{\infty}\left(\prod_{k=1}^{\fbox{{\bf n-1}}}\cos\left(\frac{k \pi}{n}\right)\right)$, we know that $z^{n-1}+z^{n-2}+\ldots+z+1=\prod_{k=1}^{n-1}\left(z-e^{\frac{2k\pi i }{n}}\right)\;\overset{z=-1}{\Rightarrow}\;\overbrace{(-1)^{n-1}+(-1)^{n-2}+\ldots+-1+1}^{\Large\frac{1}{2}\left(1+(-1)^{n}\right)}=\prod_{k=1}^{n-1}\left(-1-e^{\frac{2k\pi i }{n}\right)\Rightarrow$ $\prod_{k=1}^{n-1}\left(1+e^{\frac{2k\pi i }{n}}\right)=-\frac{1}{2}\left(1-(-1)^{n}\right)\Rightarrow \prod_{k=1}^{n-1}e^{\frac{k\pi i }{n}}\left(e^{-\frac{k \pi i }{n}}+e^{\frac{k\pi i }{n}}\right)=-\frac{1}{2}\left(1+(-1)^{n}\right)\Rightarrow\\ e^{\frac{\pi i }{n}\left(1+2+3+\ldots+\left(n-1\right)\right)}\prod_{k=1}^{n-1}2\cos\left(\frac{k\pi}{n}\right)=-\frac{1}{2}\left(1+(-1)^{n}\right)\Rightarrow e^{\frac{i\pi(n-1)}{2}}\prod_{k=1}^{n-1}2\cos\left(\frac{k\pi}{n}\right)=-\frac{1}{2}\left(1+(-1)^{n}\right)\Rightarrow$ $i^{n-1}2^{n-1}\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)=-\frac{1}{2}\left(1-(-1)^{n}\right)\Rightarrow\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)=-\frac{1}{i^{n-1}}\frac{1}{2^{n}}\left(1-(-1)^{n}\right)=\begin{cases}0&,n=2k\\\frac{(-1)^{\frac{n-1}{2}}}{2^{n-1}}&,n=2k+1\end{cases}$ so $\sum_{n=1}^{\infty}\left(\prod_{k=1}^{n-1}\cos\left(\frac{k \pi}{n}\right)\right)=-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2^{2n}}=-\sum_{n=0}^{\infty}\left(-\frac{1}{4}\right)^{n}=-\frac{1}{1+\frac{1}{4}}=-\frac{4}{5}$.[/color]  July 20th, 2012, 03:46 AM #7 Senior Member Joined: May 2011 Posts: 501 Thanks: 6 Re: Product-Series Computation .... deleted. Unnecessary. Sorry. July 20th, 2012, 04:36 AM #8 Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Product-Series Computation [color=#000000]Hi G., Mark already gave these formulas in your post in the following post.[/color] Quote:  Originally Posted by MarkFL $\sum_{n=1}^{\infty}\prod_{k=1}^{n}\cos\(\frac{k\pi }{n}$$=\sum_{n=1}^{\infty}$$-\frac{\sin\(\frac{\pi}{2}n$$}{2^{n-1}}\)=4\sum_{n=1}^{\infty}$$-\frac{1}{4}$$^n=-\frac{4}{5}$ I was aided largely by the identity: $\prod_{k=1}^{n-1}\cos$$\frac{k\pi}{n}$$=\frac{\sin\(\frac{\pi}{2} n)}{2^{n-1}}$

 July 20th, 2012, 08:23 AM #9 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Product-Series Computation Okey-doke. I am sorry. I did not notice.

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