Question

Find the sum of the series \(\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^{2}}\right)\).

Short answer

The sum of the series \(\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^{2}})\) is \(\ln(4)\)

Step-by-step solution

Identify the series

The series given is \(\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^{2}})\)

Rewrite the series

Redefine the series using logarithm properties. As \(\ln(ax) = \ln(a) + \ln(x)\), and \(\ln(1-\frac{1}{n^{2}}) = \ln( \frac{n^{2}-1}{n^{2}} ) = \ln(n^{2}) - \ln(n^{2}-1)\). So the series now becomes \(\sum_{n=2}^{\infty}[\ln(n^{2}) - \ln(n^{2}-1)]\).

Split the series

Split the series into two, with the general terms \(\ln(n^{2})\) and \(- \ln(n^{2}-1)\). So, the series can be written as \(\sum_{n=2}^{\infty}\ln(n^{2}) - \sum_{n=2}^{\infty}\ln(n^{2}-1)\)

Identify the telescoping series

A telescoping series is formed in \(\sum_{n=2}^{\infty}\ln(n^{2}) - \sum_{n=2}^{\infty}\ln(n^{2}-1)\). Here, each term in \(\sum_{n=2}^{\infty}\ln(n^{2}-1)\) cancels out with the corresponding term in \(\sum_{n=2}^{\infty}\ln(n^{2})\). Specifically, the term in the first series when n=k cancels out with the term in the second series when n=k+1.

Find the sum of the series

After the cancellation in the telescoping series, the remaining terms are \(\ln(4) - \ln(1) = \ln(4)\). As log to the base e of 1 is 0, the sum of the series is \(\ln(4)\).