Question: Calculate the sum of series:

$$\sum\_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$

I tried to spread this logarithm, but I'm not seeing any method for this exercise.

Answer: 
Note
$$\ln\left(\frac{n(n+2)}{(n+1)^2}\right)=\ln\left(\frac{\frac{n}{n+1}}{\frac{n+1}
{n+2}}\right)=\ln\left(\frac{n}{n+1}\right)-\ln\left(\frac{n+1}{n+2}\right)$$ 
Thus
$$\sum\_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)=\sum\_{n=1}^\infty \left[\ln\left(\frac{n}{n+1}\right)-\ln\left(\frac{n+1}{n+2}\right)\right]$$
This is a [telescoping series](https://en.wikipedia.org/wiki/Telescoping_series). Therefore
$$\sum\_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)=-\ln(2)$$
\begin{align\*}\sum\_{n=1}^N \ln\left(\frac{n(n+2)}{(n+1)^2}\right)&=
\sum\_{n=1}^N \left(\ln n + \ln (n+2) -2\ln(n+1)\right)\\&=\ln 1+\ln(N+2)-\ln2-\ln(N+1)\\&=-\ln2+\ln\frac{N+2}{N+1}\\&\xrightarrow{N\to\infty}-\ln2+\ln1=-\ln2.\end{align\*}
This is however essentially the same solution as that given by Behrouz, just less clever and explicit about the limit.