Evaluating the following summation $\require{cancel}$
Evaluate $$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right)$$
My procedure:
$$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right)=\sum_{n=2}^{\infty}(\ln\left(n-1\right)-\ln(n)+\ln(n+1)-\ln(n))$$
If we evaluate a few terms:
$$\sum_{n=2}^{m}(\ln\left(n-1\right)-\ln(n)+\ln(n+1)-\ln(n))= $$
$$=0-\ln(2)\cancel{+\ln(3)}\cancel{-\ln(2)+\ln(2)}\cancel{-\ln(3)}\cancel{+\ln(4)} \cancel{-\ln(3)+\ln(3)}\cancel{-\ln(4)}+\ln(5)-\ln(4)+...+\ln(m+1)-\ln(m)$$
Thus: $$\lim_{m\to\infty}\sum_{n=2}^{m}\ln\left(\frac{n^2-1}{n^2}\right)=\lim_{m\to\infty}\left(-\ln(2)+\ln(m+1)-\ln(m)\right)=\boxed{-\ln(2)}$$
However, I am not really happy with this method. Is there any other way? A more elegant one? 
 A: Step by step telescoping below:
$$\require{cancel}
\begin{align}
& \sum_{n=2}^{\infty}\big(\ln\left(n-1\right)-2 \ln(n)+\ln(n+1)\big) \\
 &\;\;= \sum_{n=2}^{\infty}\big(\left(\ln(n+1)-\ln(n)\right)-\left(\ln(n)-\ln\left(n-1\right)\right)\big) \\
 &\;\;= \sum_{n=2}^{\infty}\big(\ln(n+1)-\ln(n)\big)- \sum_{n=2}^{\infty}\big(  \ln(n)-\ln\left(n-1\right)\big) \\
 &\;\;=  \sum_{\color{red}{n=3}}^{\infty}\big(\ln(n)-\ln(n-1)\big)- \sum_{n=2}^{\infty}\big(  \ln(n)-\ln\left(n-1\right)\big) \\
 &\;\;= \bcancel{\sum_{n=3}^{\infty}\big(\ln(n)-\ln(n-1)\big)}-\big(\ln(2)-\ln(2-1)\big)- \bcancel{\sum_{\color{red}{n=3}}^{\infty}\big(  \ln(n)-\ln\left(n-1\right)\big)} \\[3px]
 &\;\;= -\ln(2)
\end{align}
$$
A: 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 $-\ln 2$. 
