What is $\lim_{n\rightarrow \infty} \frac{1}{n^2}(\ln(\frac{2^n}{3^n})+\ln(\frac{5^n}{4^n})+\cdots+\ln(\frac{(3n-1)^n}{(n+2)^n}))$? Per the title of this question, how does one go about calculating $$\lim_{n\rightarrow \infty} \frac{1}{n^2}\left(\ln\left(\frac{2^n}{3^n}\right)+\ln\left(\frac{5^n}{4^n}\right)+\cdots+\ln\left(\frac{(3n-1)^n}{(n+2)^n}\right)\right)\  ?$$
Thanks!

For a mechanical* way to do this:
Hint:
$$\log \frac{(3k-1)}{k+2} = \log 3 + \log (1- \frac{1}{3k}) - \log (1 + \frac{2}{k})$$
and 
$$\log (1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots $$
for $|x| \lt 1$.
More details:
Using the above hint, the $k^{th}$ term is 
 $$ n \log \frac{(3k-1)}{k+2} = n \log 3 + \frac{5n}{3k} + \mathcal{O}(\frac{n}{k^2})$$
and so your sum is
$$ \frac{1}{n^2} \sum_{k=1}^{n} (n \log 3 + \frac{5n}{3k} + \mathcal{O}(\frac{n}{k^2}))$$
$$ = \frac{1}{n^2}(n^2\log 3 + \frac{5n\log n}{3} + \mathcal{O}(n))$$
Here we used the estimate $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \log n + \mathcal{O}(1)$ and $\displaystyle \sum_{k=1}^{n} \frac{1}{k^2} = \mathcal{O}(1)$
Thus the sum is 
  $$ = \log 3 + \frac{5\log n}{3n} + \mathcal{O}(\frac{1}{n})$$
and so your limit is
 $$\log 3$$
Note that we don't really need the estimate $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \log n + \mathcal{O}(1)$
All we need to show is that $\displaystyle \sum_{k=1}^{n} \frac{1}{k}  = o(n)$ and this easily follows from the following classic theorem:
If $\displaystyle a_n \to 0$, then $\displaystyle \frac{1}{n} \sum_{k=1}^{n} a_k \to 0$
*As Didier points out (see comments below), this last theorem can be used to skip all the mechanical calculations done above by applying it directly to the terms of your sequence.
