A Natural Sum

$Log\left( \left( \frac { 1 }{ 2 } \frac { 3 }{ 2 } \right) \left( \frac { 2 }{ 3 } \frac { 4 }{ 3 } \right) \left( \frac { 3 }{ 4 } \frac { 5 }{ 4 } \right) ... \right) =Log\left( \frac { 1 }{ 2 } \right)$

$\ln{(1-\frac{1}{n^2})}=\ln{(\frac{n^2-1}{n^2})}=\ln{\frac{(n+1)(n-1)}{n^2}}=\ln{[(n+1)(n-1)]}-\ln{n^2}$

$=\ln{(n+1)}+\ln{(n-1)}-2\ln{n}=\ln{(n-1)}-\ln{n}-\ln{n}+\ln{(n+1)}$

$=\ln{\frac{n-1}{n}}-[\ln{n}-\ln{(n+1)}]=\ln{\frac{n-1}{n}}-\ln{\frac{n}{n+1}}$ .

Let $s_k=\sum _{ n=2}^{k }{\ln{(1-\frac{1}{n^2})}}=\sum _{ n=2}^{k}{(\ln{\frac{n-1}{n}}}-\ln{\frac{n}{n+1}})$ for $k \ge 2$ .

Then

$s_k=(\ln{\frac{1}{2}}-\ln{\frac{2}{3}})+(\ln{\frac{2}{3}}-\ln{\frac{3}{4}})+...+(\ln{\frac{k-1}{k}}-\ln{\frac{k}{k+1}})=\ln{\frac{1}{2}}-\ln{\frac{k}{k+1}},$ so

$\sum _{ n=2}^{\infty}{\ln{(1-\frac{1}{n^2})}}=\lim_{k\rightarrow \infty}{(\ln{\frac{1}{2}}-\ln{\frac{k}{k+1}})}=\ln{\frac{1}{2}}-\ln{1}=\ln{1}-\ln{2}-\ln{1}=\boxed{-\ln{2}}.$

$\displaystyle \sum_{n=2}^{\infty} \ln (1-\frac{1}{n^{2}})=\displaystyle \sum_{n=2}^{\infty} \ln (\frac{(n+1)}{n} \cdot \frac{(n-1)}{n})$ $=\displaystyle \sum_{n=2}^{\infty} (\ln (n+1) - \ln (n)) + \displaystyle \sum_{n=2}^{\infty} (\ln (n-1) - \ln (n))$ $=\displaystyle \lim_{n\to\infty} (\ln (n+1) - ln (2)) + \lim_{n\to\infty} (\ln (1) - \ln (n))$ $=\ln (\lim_{n\to\infty} \frac{n+1}{n}) - \ln (2)$ $=\ln(1) - \ln (2)=-\ln (2)$