Show the limit of the following is $\dfrac{1}{12}$ 
Show that $$\lim_{n\to \infty}n^2 \log \left(\dfrac{r_{n+1}}{r_{n}} \right)=\dfrac{1}{12}$$ where $r_{n}$ is defined as;
  $$r_{n}=\dfrac{\sqrt{n}}{n!} \left(\dfrac{n}{e} \right)^n$$.  

Now I simplified $\dfrac{r_{n+1}}{r_{n}}$ to $$\dfrac{r_{n+1}}{r_{n}}= \left(\dfrac{n+1}{n} \right)^{n+0.5} \dfrac{1}{e}$$
And so simplified $n^{2} \log \left(\dfrac{r_{n+1}}{r_{n}} \right)$ to $$n^{2}\left((n+0.5) \log \left(\dfrac{n+1}{n} \right)-1 \right).$$
Now I checked this on wolfram and as $n \rightarrow \infty$ this does converge t $\dfrac{1}{12}$ but I am not able to show it.
I have tried L'Hopital i.e $$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}}$$
Now as $n \rightarrow \infty$  $\log \left(\dfrac{n+1}{n} \right) \rightarrow 0$ but you end up with a $\dfrac{-1}{0}$ type limit.
Any help with this would be much appreciated.  
EDIT: I have realised I have made a mistake and that $$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}}$$  gives you a $\dfrac{0}{0}$ type limit, will delve into the algebra and get back.

Application of the rule of L'Hospital is a good idea:
$$\lim_{n\to \infty} \dfrac{n \log \left(\dfrac{n+1}{n} \right) +0.5 \log \left(\dfrac{n+1}{n} \right)-1}{\dfrac{1}{n^2}} =\lim_{x\to 0} \dfrac{\dfrac{1}{x} \log(1+x) +0.5 \log (1+x)-1}{x^2} =(L'H) \lim_{x\to 0} \dfrac{\dfrac{-1}{x^2} \log(1+x)+\dfrac{1}{x(1+x)}+\dfrac{1}{2(1+x)} }{2x} =(L'H)\lim_{x\to 0} \dfrac{- \log(1+x)+\dfrac{x(x+2)}{2(1+x)} }{2x^3} =\lim_{x\to 0} \dfrac{- \dfrac{1}{1+x}+\dfrac{x^2+2x+2}{2(1+x)^2} }{6x^2} =\lim_{x\to 0} \dfrac{1}{12(1+x)^2}= \dfrac{1}{12}$$
