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).