I´ve got the following infinite sum which - determined by the ratio test - is convergent for $\{p\in\mathbb{R}\ |\ p>1\}$
$$\left(\frac{1}{2\sqrt{\pi}}\right)\cdot\sum\limits_{n=1}^\infty \frac{(2p-1)^{n-1}}{p^{2n-1}}\cdot\frac{1}{\sqrt{n-1}}.$$
This series is a result of applying Stirling's approximation to the sum of an infinite series given by (that's why the whole sum is multiplied by a constant):
$$\frac{1}{2}\cdot\sum\limits_{n=1}^\infty \frac{(2p-1)^{n-1}}{p}\cdot\frac{1}{p^{2n-2}}\cdot\frac{(2n-2)!}{(n-1)!(n-1)!}$$
Which can be expanded to:
$$2\sum\limits_{n=1}^\infty \frac{(2p-1)^{n-1}}{p}\cdot\frac{1}{p^{2n-2}}\cdot\left(\frac{(2n-3)}{(2n-2)}\cdot\frac{(2n-5)}{(2n-4)}\cdot\frac{(2n-7)}{(2n-6)}\cdot\frac{(2n-9)}{(2n-8)}\ \cdots\ \frac{3}{2}\right)$$
My question is the following: how would I go about approximating - if it is even possible to find the exact solution - either this series or its Stirling's approximate? And assuming it's impossible: if I was to compute the result, how would I show that the result is correct to a given number of decimal places?
Stirling's approximation only applies for large $n$: it certainly doesn't apply for $n-1=0$!
Actually, as written, I don't think you need to take Stirling's approximation for your sum: the coefficients are the central binomial coefficients, $\binom{2n}{n}$; these have generating function $$ \sum_{n=0}^{\infty} \binom{2n}{n} z^n = \frac{1}{\sqrt{1-4z}}, $$ and rewriting your sum, I think it becomes $$ \frac{1}{2p} \sum_{n=0}^{\infty} \left( \frac{2p-1}{p^2} \right)^n \binom{2n}{n} = \frac{1}{2p} \left( 1-4\frac{2p-1}{p^2} \right)^{-1/2} = \frac{1}{2\sqrt{p^2-8p+4}} $$ when $p>0$. However, we also need $z=\frac{2p-1}{p^2}<\frac{1}{4}$, which reduces to $p>2(2+\sqrt{3})$ or $p<2(2-\sqrt{3})$.