Homework Help: Math

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 =

(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.
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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