Wolfram Alpha can't calculate this limit, can you?

Hi Sir, You might wish to learn a bit of latex here

Amazingly, I didn't know that the problem was worth a few lines. The solution I thought of was this:

First , we expand both numerator and denominator to get :

$\frac{p}{q} = \displaystyle \lim_{n \to \infty} \displaystyle \bigg(\prod_{r=1}^{2n} \bigg(\frac{3n+r}{n+r}\bigg)\bigg)^{\frac{1}{n}}$

$\ln \bigg(\frac{p}{q}\bigg) = \displaystyle \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \ln \bigg(\frac{3n+r}{n+r}\bigg)$

$\Rightarrow \ln \bigg(\frac{p}{q}\bigg) = \displaystyle \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \ln \bigg(\frac{3+r/n}{1+r/n}\bigg)$

$\Rightarrow \ln \bigg(\frac{p}{q}\bigg) = \displaystyle \int_{0}^{2} \ln \bigg(\frac{3+x}{1+x}\bigg)dx = \ln\bigg(\frac{3125}{729}\bigg)$

$\Rightarrow \frac{p}{q} = \frac{3125}{729}$

Is this Sterling's approximation? If so, please tell how do we get the answer from it ?

n! can be approximated by √(2πn) (n/e)^n. I thought that was the easiest way to work this one out. Your solution is very interesting. Always great to look at different ways of solving the same problem.