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.

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.

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]

Ah, thanks! Makes perfect sense

To determine the convergence of the given series Σ ln (1 - 1/k^2), let's start by rewriting it as Σ ln( (k^2 - 1) / k^2).

Now, we can simplify the term inside the natural logarithm using the property ln(a/b) = ln(a) - ln(b):

ln( (k^2 - 1) / k^2) = ln(k^2 - 1) - ln(k^2).

From here, we want to transform the sum into a product (as suggested in the assumption). However, it's crucial to note that this series starts at k = 2. Therefore, we need to adjust the expression accordingly:

Π (1 - 1/k^2) = Π (1 - 1/k^2) * ln(1/1) * ln(2/2).

Now, we have:

Π ln( (k^2 - 1) / k^2) = Π ln(k^2 - 1) - Π ln(k^2).

To calculate the product of these logarithms, we can evaluate each term separately and then take the difference of the corresponding products:

For the first product:

Π ln(k^2 - 1) = ln(1) + ln(3^2 - 1) + ln(4^2 - 1) + ln(5^2 - 1) + ...

And for the second product:

Π ln(k^2) = ln(2^2) + ln(3^2) + ln(4^2) + ln(5^2) + ...

Now, we can see that Π ln( (k^2 - 1) / k^2) is equivalent to the difference between these two products. By analyzing the terms, we observe a cancellation pattern:

ln(1) in the first product and ln(2^2) in the second product.
ln(3^2 - 1) in the first product and ln(3^2) in the second product.
ln(4^2 - 1) in the first product and ln(4^2) in the second product.
...

In general, for every term ln((k^2 - 1) / k^2) in the first product, we have ln(k^2) in the second product, except for the initial term.

Therefore, all the intermediate terms cancel out, leaving us with only the ln(1) term in the first product and the ln(2^2) term in the second product.

Thus, the series can be simplified to:

ln(1) - ln(2^2) = 0 - ln(2^2) = -ln(4).

Hence, the given infinite series converges to -ln(4).