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:
11/k^2 = (k^21)/k^2 =
(k+1)(k1)/k^2
We can write this as:
f(k+1)/f(k)
where
f(k) = k/(k1)
So, then we ave:
f(k+1)/f(k)=
(k+1)/k * (k1)/k
which is exactly the kth term.
The product can then be written as:
[f(3)/f(2)]*[f(4)/f(3)]*[f(5)/f(4)]*...
= 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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