# How does this limit of a sum work?

So: I'm trying to work out how this argument works. I think it's using a Riemann sum to take the limit but I never took real analysis so, I'm not familiar with how to actually work with limits and sums like this. I don't strictly speaking need to know why it works- the result is enough- but I'd like to.

$\overset{\sim}{W}_n$ is a random variable which takes integer values from $0$ to $n$.

$\textbf{P}\left(\overset{\sim}{W}_n=k\right)= \frac{\Gamma\left(k+W_0\right)}{\Gamma\left(k+1\right)} \frac{\Gamma\left(n-k+B_0\right)}{\Gamma\left(n-k+1\right)} \frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\tau_0\right)} \frac{\Gamma\left(\tau_0\right)}{\Gamma\left(W_0\right)\Gamma(B_0)}$

For $x\in \left[0,1\right]$:

$\textbf{P}\left(\overset{\sim}{W}_n\leq nx\right)= \sum\limits_{k=0}^{\lfloor nx\rfloor} \frac{\Gamma\left(k+W_0\right)}{\Gamma\left(k+1\right)} \frac{\Gamma\left(n-k+B_0\right)}{\Gamma\left(n-k+1\right)} \frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\tau_0\right)} \frac{\Gamma\left(\tau_0\right)}{\Gamma\left(W_0\right)\Gamma(B_0)}$

Using Stirling's approximation to the ratio of gamma functions: $\frac{\Gamma\left(x+r\right)}{\Gamma\left(x+s\right)} = x^{r-s} + O\left(x^{r-s-1}\right) \text{ as x$\rightarrow\infty$}$

and taking the limit as $n\rightarrow\infty$:

Aka, a miracle occurs. How do we get the following result?

$\textbf{P}\left(\frac{\overset{\sim}{W}_n}{n}\leq x\right)= \frac{\Gamma\left(\tau_0\right)}{\Gamma\left(W_0\right)\Gamma(B_0)} \int\limits_{0}^{x} u^{W_0-1}\left(1-u\right)^{B_0-1} du$

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