The sum of series with natural logarithm: $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ 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.
 A: In another, more straight, way:
$$
\begin{gathered}
  \sum\limits_{1\, \leqslant \,n} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)}  = \ln \left( {\prod\limits_{1\, \leqslant \,n} {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} } \right) =  \hfill \\
   = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n} n }}
{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 1} \right)} }}\;\frac{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 2} \right)} }}
{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 1} \right)} }}} \right) = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n} n }}
{{\prod\limits_{2\, \leqslant \,n} n }}\;\frac{{\prod\limits_{3\, \leqslant \,n} n }}
{{\prod\limits_{2\, \leqslant \,n} n }}} \right) =  \hfill \\
   = \ln \left( {1\;\frac{1}
{2}} \right) = \ln \left( {\frac{1}
{2}} \right) \hfill \\ 
\end{gathered} 
$$
Rewriting the above in  more rigorous terms, we have
$$
\begin{gathered}
  \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)}  = \ln \left( {\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} } \right) =  \hfill \\
   = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} n }}
{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 1} \right)} }}\;\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 2} \right)} }}
{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 1} \right)} }}} \right) = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} n }}
{{\prod\limits_{2\, \leqslant \,n\, \leqslant \,q + 1} n }}\;\frac{{\prod\limits_{3\, \leqslant \,n\, \leqslant \,q + 2} n }}
{{\prod\limits_{2\, \leqslant \,n\, \leqslant \,q + 1} n }}} \right) =  \hfill \\
   = \ln \left( {\frac{1}
{{q + 1}}\;\frac{{q + 2}}
{2}} \right) \hfill \\ 
\end{gathered} 
$$
and therefore
$$
\mathop {\lim }\limits_{q\, \to \,\infty } \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)}  = \mathop {\lim }\limits_{q\, \to \,\infty } \ln \left( {\frac{1}
{2}\;\frac{{q + 2}}
{{q + 1}}} \right) = \ln \left( {\frac{1}
{2}\mathop {\lim }\limits_{q\, \to \,\infty } \;\frac{{q + 2}}
{{q + 1}}} \right) = \ln \left( {\frac{1}
{2}} \right)
$$
A: 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. 

For why the infinite product is $\frac {1}{2} $, see here.
A: Another overkill. Since:
$$ \frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)\tag{1} $$
we have:
$$ \sum_{n\geq 1}\log\frac{n(n+2)}{(n+1)^2}=\log\prod_{n\geq 2}\left(1-\frac{1}{n^2}\right)=\log\lim_{x\to 1}\frac{\sin(\pi x)}{\pi x(1-x^2)}\stackrel{dH}{=}\color{red}{-\log 2}.\tag{2} $$
A: \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.
A: 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. Therefore
$$\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)=-\ln(2)$$
A: 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, we have $$\sum_{n\geq1}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq0}\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+2\right)}\right)=\log\left(\frac{\Gamma\left(2\right)\Gamma\left(2\right)}{\Gamma\left(1\right)\Gamma\left(3\right)}\right)=\color{red}{\log\left(\frac{1}{2}\right)}.$$
