Nootropic Posted January 31, 2007 Report Share Posted January 31, 2007 So I ran into a question today involving an infinite logarithmic series which asked to how to show that ln(1-(1/k^2)) (the sum of this infinite series) is -ln 2. To make it short, I have absolutely no idea how to... My first idea was to write a closed form, but I'm really not quite sure. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted February 1, 2007 Report Share Posted February 1, 2007 so you mean this? Prove: [math]\sum^{\infty}_{k=2}Log_e(1-\frac{1}{k^2})=-Log_e(2)[/math] Quote Link to comment Share on other sites More sharing options...
Nootropic Posted February 1, 2007 Author Report Share Posted February 1, 2007 Yes, except it would be the natural logarithm. Excuse the latex deficiencies. Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 2, 2007 Report Share Posted February 2, 2007 I think the equation should be [math]\sum^{\infty}_{k=2}\log(1-\frac{1}{k^2})=-\log(2)[/math] Because [math]\lim_{a \rightarrow 1} \log(1-\frac{1}{a^2}) = \infty [/math], the first term of the series [math]\sum^{\infty}_{k=1}\log(1-\frac{1}{k^2})[/math] is undefined. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted February 2, 2007 Report Share Posted February 2, 2007 right you are craig :) off to get last years books and see if i actually learnt this, because I dont remember it! Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 2, 2007 Report Share Posted February 2, 2007 I'd say it's equivalent to proving: [math]\prod_{k>1}(1-\frac{1}{k^2})=\frac12[/math], which looks vaguely reminescent of the definition of e........ BTW, the LaTeX command for natural logarithm is \ln. Quote Link to comment Share on other sites More sharing options...
Nootropic Posted February 7, 2007 Author Report Share Posted February 7, 2007 does anyone know? There seems to be nothing online about logarithmic series and my book doesn't help, and my teacher doesn't really know, so I'm pretty much out of resources other than my uniformed brain. Quote Link to comment Share on other sites More sharing options...
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