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: 
An overkill. Since holds $$\prod\_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d
 $$ and this can be proved using the [Euler's definition of the Gamma function](https://en.wikipedia.org/wiki/Gamma_function#Alternative_definitions), we have $$\sum\_{n\geq1}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod\_{n\geq0}\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+2\right)}\right)=\log\left(\frac{\Gamma\left(2\right)\Gamma\left(2\right)}{\Gamma\left(1\right)\Gamma\left(3\right)}\right)=\color{red}{\log\left(\frac{1}{2}\right)}.$$
We can see that $$\frac {n (n+2)}{(n+1)^2} =1-\frac {1}{(n+1)^2} $$ So we have, $$\sum\_{n=1}^{\infty} \log \left(1-\frac {1}{(n+1)^2}\right) =\lim\_{n \to \infty} \log \left(\left(1-\frac {1}{4}\right)\left(1-\frac {1}{9}\right)\cdots\left(1-\frac {1}{(n+1)^2}\right)\right) =\log \frac {1}{2} $$ Hope it helps. 

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For why the infinite product is $\frac {1}{2} $, see [here](https://math.stackexchange.com/questions/18179/finding-value-of-the-infinite-product-prod-bigl1-frac1n2-bigr).