How do i prove this? 
Bartle - Introduction to Analysis p.275
Define $x_n = \frac{a(a+1)\cdots(a+n)b(b+1)\cdots(b+n)}{n!c(c+1)\cdots(c+n)}$.
Show there $\sum x_n$ converges if $c>a+b$ and diverges if $c<a+b$.

How do i prove this?
I was trying to apply Raabe's Test, so i found that $\lim n(|\frac{x_{n+1}}{x_n}| - 1) = a+b-c$. However i think this does not answer the question.
How do i prove this?
 A: You have
$$\frac{x_n}{x_{n-1}}=\frac{n^2+(a+b)n+ab}{n^2+cn},\tag{1}$$
hence if $a+b\geq c$ you have:
$$ \frac{x_n}{x_{n-1}}\geq 1+\frac{ab}{n^2+cn}, \tag{2}$$
so:
$$ x_n \geq x_0\prod_{k=1}^{n}\left(1+\frac{ab}{n^2+cn}\right),\tag{3}$$
and since the product on the right is convergent $x_n$ cannot be infinitesimal, hence $\sum x_n$ cannot converge. On the other hand, if $a+b<c$ then $d=c-(a+b)>0$ and:
$$ \frac{x_n}{x_{n-1}}=1+\frac{ab-dn}{n^2+cn}\ll\left(1-\frac{1}{n}\right)^d, $$
hence:
$$ x_n \ll \frac{1}{n^{d}}$$
and, if $c>a+b+1$, the series $\sum x_n$ converges. Since the $\ll$-symbol in the above lines can be replaced by the $\sim$-symbol, I believe that the correct statement is:

Given $x_n = \frac{(a)_{n+1}(b)_{n+1}}{n!\cdot(c)_{n+1}}$, $\sum x_n$
  converges if $c>a+b+1$ and diverges if $c<a+b+1$.

As a matter of fact, if we take $a=b=1$ and $c=\frac{5}{2}$, we have:
$$ x_n = \frac{15\sqrt{\pi}}{8}\cdot \frac{(n+1)\Gamma(n+2)}{\Gamma(n+7/2)}\sim\frac{1}{\sqrt{n}},$$
hence $\sum x_n$ does not converge as expected by your claim.
