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.

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
$$
\mathop {\lim }\limits_{q\, \to \,\infty } \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}}
{{\left( {n + 1} \right)^{\,2} }}} \right)}  = \mathop {\lim }\limits_{q\, \to \,\infty } \ln \left( {\frac{1}
{2}\;\frac{{q + 2}}
{{q + 1}}} \right) = \ln \left( {\frac{1}
{2}\mathop {\lim }\limits_{q\, \to \,\infty } \;\frac{{q + 2}}
{{q + 1}}} \right) = \ln \left( {\frac{1}
{2}} \right)
$$
