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