The kth factor in the product is:
1-1/k^2 = (k^2-1)/k^2 =
(k+1)(k-1)/k^2
We can write this as:
f(k+1)/f(k)
where
f(k) = k/(k-1)
So, then we ave:
f(k+1)/f(k)=
(k+1)/k * (k-1)/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.
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.
3 answers
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]
sin(pi x)/(pi x) =
Product from k = 1 to infinity of
[1 - x^2/k^2]
Ah, thanks! Makes perfect sense