Show that $\sum\limits_{k=1}^{\infty}\ln\left(\frac{k(k+2)}{(k+1)^2}\right) = -\ln(2)$? I'm working with the following sum and trying to determine what it converges to: $$\sum_{k=1}^{\infty}\ln\left(\frac{k(k+2)}{(k+1)^2}\right)$$
Numerically I see that it seems to be converging to $-\ln(2)$, however I can't see why that is the case. I have expanded the logarithm and expressed the sum as $\sum_{k=1}^{\infty}\ln(k)+\ln(k+2)-2\ln(k+1)$, but I don't know if that actually helps anything. Does anyone have any thoughts on how to approach this problem?
 A: $$\frac{k^2+2k}{k^2+2k+1}=1-\frac1{(k+1)^2}\implies$$
$$\sum_{k=1}^n\log\left(1-\frac1{(k+1)^2}\right)=\sum_{k=1}^n\log\left[\left(1-\frac1{k+1}\right)\left(1+\frac1{k+1}\right)\right]=$$
$$=\sum_{k=1}^n\left[\log\left(1-\frac1{k+1}\right)+\log\left(1+\frac1{k+1}\right)\right]=$$
$$=\log\frac12+\overbrace{\log\frac32+\log\frac23}^{=\log1=0}+\overbrace{\log\frac43+\log\frac34}^{=\log1=0}+\log\frac54+\ldots+\log\frac n{n+1}+\log\frac{n+2}{n+1}=$$
$$=\log\frac12+\log\frac{n+2}{n+1}\xrightarrow[n\to\infty]{}-\log2$$
A: In another way:
$$
\eqalign{
  & \sum\limits_{k\, = \,1}^n {\log \left( {{{k\left( {k + 2} \right)} \over {\left( {k + 1} \right)^2 }}} \right)}  = \log \prod\limits_{k\, = \,1}^n {{{k\left( {k + 2} \right)} \over {\left( {k + 1} \right)^2 }}}  =   \cr 
  &  = \log \left( {\prod\limits_{k\, = \,1}^n {{k \over {\left( {k + 1} \right)}}} \;\prod\limits_{k\, = \,1}^n {{{\left( {k + 2} \right)} \over {\left( {k + 1} \right)}}} } \right) = ({\rm telescoping})  \cr 
  &  = \log \left( {{1 \over {\left( {n + 1} \right)}}\;{{n + 2} \over 2}} \right) \cr} 
$$
A: $$
\begin{align}
\sum_{k=1}^n\log\left(\frac{k(k+2)}{(k+1)^2}\right)
&=\sum_{k=1}^n\log(k)+\sum_{k=1}^n\log(k+2)-2\sum_{k=1}^n\log(k+1)\\
&=\sum_{k=1}^n\log(k)+\sum_{k=3}^{n+2}\log(k)-2\sum_{k=2}^{n+1}\log(k)\\[3pt]
&=\log(1)-\log(2)+\log(n+2)-\log(n+1)\\[6pt]
&=-\log(2)+\log\left(\frac{n+2}{n+1}\right)
\end{align}
$$
Let $n\to\infty$,
$$
\sum_{k=1}^\infty\log\left(\frac{k(k+2)}{(k+1)^2}\right)=-\log(2)
$$
