Posted by Sean on Thursday, May 28, 2009 at 12:28pm.
What does the following infinite series starting at k=2 converge to: Σ ln (1 - 1/k^2)
In other words, what does this converge to: ln(1 - 1/4) + ln(1 - 1/9) + ln(1 - 1/16) + ln(1 - 1/25) + ln(1 - 1/36) + ...
I assume the first step is this:
Σ ln (1 - 1/k^2) = ln Π (1 - 1/k^2)
But from there, I don't know how to convert this to closed form and continue.
- Math - Count Iblis, Thursday, May 28, 2009 at 12:59pm
The kth factor in the product is:
1-1/k^2 = (k^2-1)/k^2 =
We can write this as:
f(k) = k/(k-1)
So, then we ave:
(k+1)/k * (k-1)/k
which is exactly the kth term.
The product can then be written as:
= 1/f(2) as all the other factors cancel.
- Math - Count Iblis, Thursday, May 28, 2009 at 1:05pm
Note that this is a special case of the formula:
sin(pi x)/(pi x) =
Product from k = 1 to infinity of
[1 - x^2/k^2]
- Math - Sean, Thursday, May 28, 2009 at 1:28pm
Ah, thanks! Makes perfect sense
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