Evaluating the following summation

Step by step telescoping below:

$$\require{cancel} \begin{align} & \sum_{n=2}^{\infty}\big(\ln\left(n-1\right)-2 \ln(n)+\ln(n+1)\big) \\ &\;\;= \sum_{n=2}^{\infty}\big(\left(\ln(n+1)-\ln(n)\right)-\left(\ln(n)-\ln\left(n-1\right)\right)\big) \\ &\;\;= \sum_{n=2}^{\infty}\big(\ln(n+1)-\ln(n)\big)- \sum_{n=2}^{\infty}\big( \ln(n)-\ln\left(n-1\right)\big) \\ &\;\;= \sum_{\color{red}{n=3}}^{\infty}\big(\ln(n)-\ln(n-1)\big)- \sum_{n=2}^{\infty}\big( \ln(n)-\ln\left(n-1\right)\big) \\ &\;\;= \bcancel{\sum_{n=3}^{\infty}\big(\ln(n)-\ln(n-1)\big)}-\big(\ln(2)-\ln(2-1)\big)- \bcancel{\sum_{\color{red}{n=3}}^{\infty}\big( \ln(n)-\ln\left(n-1\right)\big)} \\[3px] &\;\;= -\ln(2) \end{align} $$

Another way would be to consider the exponentiation. Observe \begin{align} \exp\left(\sum^N_{n=2}\ln\frac{n^2-1}{n^2} \right)=&\ \prod^N_{n=2}\frac{n^2-1}{n^2} = \prod^N_{n=2}\frac{(n-1)(n+1)}{n \cdot n}\\ =&\ \frac{(2-1)(2+1)}{2\cdot 2}\cdot \frac{(3-1)(3+1)}{3\cdot 3}\cdots\frac{(N-2)N}{(N-1)\cdot (N-1)}\frac{(N-1)(N+1)}{N\cdot N}\\ =&\ \frac{1}{2}\frac{N+1}{N} \end{align} which means \begin{align} \sum^N_{n=2}\ln \frac{n^2-1}{n^2} = \ln\frac{N+1}{2N}. \end{align} Take the limit as $N\rightarrow \infty$ yields $-\ln 2$.