Condition for convergence of the series $S = \frac{a}{b}+\frac{a(a+1)}{b(b+1)} + \frac{a(a+1)(a+2)}{b(b+1)(b+2)} + \ldots $ Let 
$$S = \frac{a}{b}+\frac{a(a+1)}{b(b+1)} + \frac{a(a+1)(a+2)}{b(b+1)(b+2)}  + \ldots $$
What is the relationship between $a$ and $b$ for $S$ to be a convergent series ?

My attempt: 
Letting $u_n$ denote the $nth$ term in the series, we have
$$ u_n = \frac{\prod_{i=0}^{i=n-1} (a+i)}{\prod_{j=0}^{j=n-1} (b+j)} $$
We have
$$\frac{u_n}{u_{n+1}} = \frac{(b+n)}{(a+n)}$$
The ratio tests do not lead to progress. Any hints on alternative methods ?
 A: Multiply and divide your sum by $\dfrac{\Gamma(a)}{\Gamma(b)}$ . You have $S=\dfrac{\Gamma(b)}{\Gamma(a)}\displaystyle\sum_{n=1}^\infty\dfrac{\Gamma(a+n)}{\Gamma(b+n)}=\dfrac{\Gamma(b)}{\Gamma(a)}\cdot\dfrac{\Gamma(a+1)}{\Gamma(b)}$ $\cdot\dfrac1{b-a-1}=\dfrac a{b-a-1}$ . Can you take it from here? :-)
A: Letting $u_n$ denote the $nth$ term in the series, we have
$$ u_n = \frac{\prod_{i=0}^{i=n-1} (a+i)}{\prod_{j=0}^{j=n-1} (b+j)} $$
Let us try to apply the ratio test.We have
$$lim_{n \to \infty} \frac{u_n}{u_{n+1}} = lim_{n \to \infty}\frac{(b+n)}{(a+n)} = 1$$.
This is not very helpful. Let us try another test, called Raabe's test. Raabe's test can be tried whenever we have a situation like above where the ratio test gives the limit result as $1$. 
Essentially, Raabe's test  finds the following limit
$$lim_{n \to \infty} \left[ n \left( \frac{u_n}{u_{n+1}} - 1 \right) \right] = L$$
If $L > 1$, the series converges. Applying this to our problem, we have 
$$lim_{n \to \infty} \left[ n \left( \frac{b+n}{a+n} - 1 \right) \right] = lim_{n \to \infty} \left[ \frac{b-a}{\frac{a}{n}+1} \right] = (b-a)$$
Therefore, the required condition on $a,b$ for the series to be convergent is $ b - a > 1$
(Thanks to Jean-Claude Arbaut for the suggestion)
