Evaluate the series $S_n = \sum_{k=1}^n\log\frac {k (k + 2)}{(k + 1)^2}$ 
I am having trouble with (a) and (b):
I evaluated (a) and I got that sun of the first three terms does not equal $\ln(5/8)$. I got that it is equal to $\ln(3/4) + \ln(2/3) + \ln(15/16)$ which is not equal to $\ln(5/8)$.
I am also having trouble with part (b). I cannot figure out how to calculate the math to cancel out the $(k+1)^2$ and turn it into $2(n  +1)$.
Any help would be amazing.
 A: This should get you started (or more).
$$
\begin{align}
\sum_{k=1}^n\log\left(\frac{k(k+2)}{(k+1)^2}\right)
&=\sum_{k=1}^n\log\left(\frac{k}{k+1}\right)-\sum_{k=1}^n\log\left(\frac{k+1}{k+2}\right)\\
&=\sum_{k=1}^n\log\left(\frac{k}{k+1}\right)-\sum_{k=2}^{n+1}\log\left(\frac{k}{k+1}\right)\\
&=\log\left(\frac{1}{2}\right)-\log\left(\frac{n+1}{n+2}\right)\\
&=\log\left(\frac{n+2}{2n+2}\right)
\end{align}
$$
A: You can also use the identities $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$, $\ln a^m = m \ln a$, and $\ln (ab) = \ln a + \ln b$ to convert the problem into a sum:
$$
  \ln\left(\frac{k(k+2)}{(k+1)^2}\right) = \ln k + \ln (k+2) - 2 \ln(k+1)
$$
For the first three terms,
$$
 \begin{align}
  \sum &= (\ln 1 + \ln 3 - 2\ln 2) + (\ln 2 + \ln 4 - 2\ln 3) + (\ln 3 + \ln 5 - 2 \ln 4) \\
       &= \ln 1 - \ln 2 -\ln 4 + \ln 5 \\
       & = (\ln 1 + \ln 5) - (\ln 2 + \ln 4) \\
       & = \ln 5 - \ln 8 = \ln\left(\frac{5}{8}\right) \,.
 \end{align}
$$
You can compute other partial sums by generalizing this approach as suggested in the previous answer.
