This is a special case, so we assume that $a>b$ and I need to figure out if the sum $\sum_{k=1}^\infty\frac{k!b^k}{(a+kb)\cdots (a+1\cdot b)}$ diverge to infinity or not. I have tried a lot of things without succes so a hint would be appreciated.
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Setting $\lambda := a/b > 1$ you can rewrite your series as $$ \sum_{k=1}^\infty \frac{k!}{(\lambda + k)\cdots (\lambda +1)} =: \sum_k a_k. $$ Since $$ n\left(\frac{a_n}{a_{n+1}} - 1\right) = n \left( \frac{\lambda+n+1}{n+1} - 1\right) = n \frac{\lambda}{n+1} \to \lambda > 1, $$ by Raabe's criterion the series is convergent.
Rigel
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And what if $b<0$? – Mark Viola Dec 26 '17 at 21:51
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Sorry, I have implicitly assumed that $b>0$. Anyway, I think that a similar approach can be used provided that $\lambda := a/b$ is not a negative integer. – Rigel Dec 27 '17 at 06:32