Infinite Logarithmic Series

[Nootropic]

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.

[Jay-qu]

so you mean this?

 

Prove:

 

[math]\sum^{\infty}_{k=2}Log_e(1-\frac{1}{k^2})=-Log_e(2)[/math]

[Nootropic]

Yes, except it would be the natural logarithm. Excuse the latex deficiencies.

[CraigD]

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.

[Jay-qu]

right you are craig :)

 

off to get last years books and see if i actually learnt this, because I dont remember it!

[Qfwfq]

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.

[Nootropic]

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.