TITLE: $\sum_1^{\infty} \frac{(p+1)(p+2)(p+3)...(p+n)}{(q+1)(q+2)(q+3)...(q+n)}$ convergence
QUESTION [7 upvotes]: I need to determine for which values of $p$ and $q$, both greater than $0$, the following series converges: $$\sum_1^{\infty} \frac{(p+1)(p+2)(p+3)...(p+n)}{(q+1)(q+2)(q+3)...(q+n)}$$
I've tried using the ratio test, comparison test, and I've also tried partial fraction decomposition but I can't get to anything. 
Could you give me a hint on how to solve this? 
Any help will be appreciated

REPLY [5 votes]: As you have probably noticed, the ratio test is useless here. The next thing to try when the ratio test fails is the Raabe-Duhame test. In our case this amounts to computing the following limit: $\lim \limits _{n \to \infty} n(\frac {q+n+1} {p+n+1} -1) = \lim \limits _{n \to \infty} \frac {(q-p)n} {p+n+1} = q-p$. Now, if $q-p < 1$ the series diverges and if $q-p>1$ the series converges. If $q-p=1$ then the Raabe-Duhamel test fails, but in this case the series becomes $\sum \frac {p+1} {p+1+n}$. Since $p \geq 0$, the series has the same behaviour as the series $\sum \frac 1 n$ which is divergent.
Therefore the series converges only if $q-p>1$.