I've been freshening up on my sum notiation, and am trying to show the following:
For $n \in \mathbb{N}$, $n \ge 2$
$$\displaystyle \sum^n_{i=2} \ln\left(\dfrac{i}{(i-1)^2}\right) = \ln n - \sum^{n-1}_{i=2} \ln i.$$
So far, I have LHS: \begin{align*} \displaystyle \sum^n_{i=2} \ln\left(\dfrac{i}{(i-1)^2}\right)&= \displaystyle \sum^{n-1}_{i=2} \ln\left(\dfrac{i}{(i-1)^2}\right)+\ln\dfrac{n}{(n-1)^2}\\ &= \displaystyle \sum^{n-1}_{i=2}[\ln i - \ln(i-1)^2]+ \ln n - \ln (n-1)^2\\ &= \displaystyle \sum^{n-1}_{i=2}\ln i-\displaystyle \sum^{n-1}_{i=2}2 \ln (i-1)+ \ln n - 2\ln(n-1)\\ &= \displaystyle \sum^{n-1}_{i=2}\ln i-\displaystyle 2\sum^{n-1}_{i=2} \ln (i-1)+ \ln n - 2\ln(n-1)\\ \end{align*}
When I go any further, I end up looping back around to my original LHS instead of moving toward the RHS. Am I missing something with the sum notation, or is it the $\ln$'s that have me going in circles?
By the properties of the natural logarithm,
$\ln\left(\frac{i}{(i-1)^2}\right)= \ln(i) - \ln((i-1)^2)=\ln(i)-2\ln(i-1)$ so that $$ \sum_{i=2}^n\ln\left(\frac{i}{(i-1)^2}\right)=\sum_{i=2}^n \ln(i) -\sum_{i=2}^n \ln((i-1)^2) $$
Note you can "reindex" the second sum by making the change of variable $j=i-1$ to get that $$ 2\sum_{i=2}^n \ln((i-1)) = 2\sum_{j=1}^{n-1}\ln(j). $$ Now observe that the name of the variable you use in the last expression is not really important so that $$ 2\sum_{j=1}^{n-1}\ln(j)= 2\sum_{i=1}^{n-1}\ln(i) $$ So putting our observations together we get that $$ \sum_{i=2}^n\ln\left(\frac{i}{(i-1)^2}\right)= \sum_{i=2}^n \ln(i) - 2\sum_{i=1}^{n-1}\ln(i) = \ln(n) + \sum_{i=2}^{n-1} \ln(i) - 2\sum_{i=2}^{n-1}\ln(i) -2\ln(1)\\=\ln(n)-2\ln(1) - \sum_{i=2}^{n-1}\ln(i) = \ln(\frac{n}{1^2}) - \sum_{i=2}^{n-1}\ln(i) = \ln(n) - \sum_{i=2}^{n-1}\ln(i). $$