General formula for the partial sum of the series $\sum_{i=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$ 
How would I find the formula for the partial sum $S_n $ of this series? $$\sum_{k=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$$

I know that $S_1=\ln(3/4) $, $S_2=\ln(2/3) $, $S_3 =\ln(5/8)$, and $S_4=\ln(3/5)$, but I don't see any pattern that would allow me to write the partial sum formula.
 A: hint: $\ln\left(\dfrac{k(k+2)}{(k+1)^2}\right) = (\ln(k) - \ln(k+1)) + (\ln(k+2) - \ln(k+1))$
A: Consider the partial sum,
$$S_{n} = \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) $$
which can be evaluated in two ways. The first being:
\begin{align}
S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) \\
&= \sum_{k=1}^{n} ( \ln(k) + \ln(k+2) - 2 \, \ln(k+1) ) \\
&= \sum_{k=2}^{n} \ln(k) + \sum_{k=3}^{n+2} \ln(k) - 2 \, \sum_{k=2}^{n+1} \ln(k) \\
&= \sum_{k=3}^{n+2} \ln(k) - \sum_{k=2}^{n+1} \ln(k) - \ln(n+1) \\
&= \ln(n+2) - \ln(n+1) - \ln(2) = \ln\left(\frac{n+2}{2 \, (n+1)} \right).
\end{align}
The second is:
\begin{align}
S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^2}\right) = \ln\left(\prod_{k=1}^{n} \frac{k(k+2)}{(k+1)^2}\right) \\
&= \ln\left(\frac{n! \, \prod_{k=3}^{n+2} k}{\prod_{k=2}^{n+1} k^2} \right) \\
&= \ln\left(\frac{n! \, (n+2)!}{2! \, ((n+1)!)^2}\right) = \ln\left(\frac{n! \, (n+2)}{2 \, (n+1)!} \right) \\
&= \ln\left(\frac{n+2}{2(n+1)}\right).
\end{align}
In both cases 
$$S_{n} = \ln\left(\frac{1 + \frac{2}{n}}{2 \, \left(1 + \frac{1}{n}\right)} \right)$$
such that as $n \to \infty$ the result becomes $- \ln(2)$ which yields
$$\sum_{k=1}^{\infty} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) = - \ln(2).$$
A: Hint
$$\sum_{k=1}^N \ln\frac{k(k+2)}{(k+1)^2} = \ln \left(\prod_{k=1}^N \frac{k(k+2)}{(k+1)^2}  \right)= \ln \left(\frac{(\prod_{k=1}^N k)(\prod_{k=1}^N (k+2))}{(\prod_{k=1}^N (k+1))^2}  \right)= \ln \left(\frac{N! \frac{(N+2)!}{2}}{((N+1)!)^2}  \right)$$
Cancel.
