Prove that $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ converges and find its sum

The series can be turned into telescopic as follow \begin{align} S_m=\sum_{n=1}^{m}\ln\frac{n(n+2)}{(n+1)^2}&=\sum_{n=1}^m(\ln{(n+2)}+\ln n-2\ln{(n+1)}) \\ &=\sum_{n=1}^m((\ln{(n+2)}-\ln(n+1))-(\ln{(n+1)}-\ln n))) \\ &=\ln{(m+2)}-\ln(m+1)-(\ln{(2)}-\ln(1)) \end{align} So $$ \lim_{m\to\infty}S_m=\lim_{m\to\infty}\left(\ln\frac{m+2}{m+1}-\ln2\right)=-\ln2 $$

The series $$\sum_{n=1}^{\infty}\ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$ converges, using the telescoping technique: $$\begin{align} \sum_{n=1}^{\infty}\ln\left(\frac{n(n+2)}{(n+1)^2}\right)&=\left[\ln(1)+\ln(3)-2\ln(2)\right]+\left[\ln(2)+\ln(4)-2\ln(3)\right]+\left[\ln(3)+\ln(5)-2\ln(4)\right]+\cdots \end{align}$$

Note that the partial sums are $$\begin{align} S_1&=\ln(1)+\ln(3)-2\ln(2)\\ S_2&=\ln(1)+\ln(3)-2\ln(2)+\ln(2)+\ln(4)-2\ln(3)\\ &=\ln(1)-\ln(2)-\ln(3)+\ln(4)\\ S_3&=S_2+\ln(3)+\ln(5)-2\ln(4)\\ &=\ln(1)-\ln(2)-\ln(4)+\ln(5) \end{align}$$ And in general, $S_n=\ln(1)-\ln(2)-\ln(n+1)+\ln(n+2)=\ln\left(\frac{n+2}{2(n+1)}\right)$ which approaches $\ln(1/2)$.