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.

Here is the remaining part -
$$\lim_{n\to \infty} n^{2}((n + 0.5)\log(1+\frac{1}{n})-1)\\\\\,\\
\lim_{n\to \infty} n^{2}((\frac{n+0.5}{n}-\frac{n+0.5}{2n^2}+\frac{n+0.5}{3n^3} - ...) -1)\\\\\,\\
\lim_{n\to \infty} n^{2}(\frac{0.5}{n}-\frac{n+0.5}{2n^2}+\frac{n+0.5}{3n^3} - ...)\\\\\,\\
\lim_{n\to \infty} (0.5n-0.5n - \frac{1}{4} + \frac{1}{3} + \frac{0.5}{3n} - ...)\\\\\,\\
 1/3 - 1/4 = 1/12
$$
