The sum of series with natural logarithm: $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ [duplicate]

Solution 1:

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

Solution 2:

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 $$\sum_{n\geq1}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq0}\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+2\right)}\right)=\log\left(\frac{\Gamma\left(2\right)\Gamma\left(2\right)}{\Gamma\left(1\right)\Gamma\left(3\right)}\right)=\color{red}{\log\left(\frac{1}{2}\right)}.$$