I got the identity also from the Polya's Urn Scheme with waiting time, in fact the identity of Catalan Numbers listed here is a particular case from the following: $$ \displaystyle \sum_{n=0}^\infty\frac{\Gamma\big(\frac{w}{d}+n\big)}{\Gamma\big(\frac{w+1}{d}+n+1\big)} = \frac{d\cdot \Gamma\big(\frac{w}{d}\big)}{\Gamma\big(\frac{w+1}{d}\big)} $$
where $w,d\in \mathbb{Z}_+$. I am again looking for a deterministic proof.
Let $$a_n=\frac{\Gamma(\tfrac wd+n)}{\Gamma(\tfrac {w+1}d+n+1)},$$ so you are seeking $\sum_{n=0}^\infty a_n$. Letting $$ b_n=d\cdot (\tfrac{w+1}d+n)\cdot a_n, $$ it is a simple calculation to verify $$ b_n-b_{n+1}=a_n. $$ Since $b_n\to 0$ as $n\to\infty$ [check this using Stirling's approximation], we conclude $$ \sum_{n=0}^\infty a_n=b_0=d\cdot (\tfrac{w+1}d)\cdot \frac{\Gamma({w\over d})}{\Gamma({w+1\over d}+1)}=d\cdot \frac{\Gamma({w\over d})}{\Gamma({w+1\over d})} $$
