How to Compute the Sum in the Stationary Distribution of a Markov Chain?

[TaPaKaH]

Suppose we have a Markov chain with stationary distributions $p_n=\frac{a}{nb+c}p_{n-1}$ for $n\in\mathbb{N}$ where $a,b$ and $c$ are some positive constants.
It follows that $p_n=p_0\prod_{i=1}^n\frac{a}{ib+c}$. Normalisation yields $1=p_0\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}$ so $p_0=\left(\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}\right)^{-1}$.

Question: how can one compute the sum in the brackets?

 

[Mandelbroth]

Suppose we have a Markov chain with stationary distributions $p_n=\frac{a}{nb+c}p_{n-1}$ for $n\in\mathbb{N}$ where $a,b$ and $c$ are some positive constants.
It follows that $p_n=p_0\prod_{i=1}^n\frac{a}{ib+c}$. Normalisation yields $1=p_0\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}$ so $p_0=\left(\sum_{n=0}^\infty\prod_{i=1}^n\frac{a}{ib+c}\right)^{-1}$.

Question: how can one compute the sum in the brackets?

Well, let's think about this. What can we do to simplify $$\prod_{1\leq i\leq n}\frac{a}{bi+c}$$ into "workable" terms? Can you give us an attempt?