summation of a infinite series i want to know if this infinite sum $\sum_{n=1}^{\infty}$ $\frac{a(a+1)(a+2)......(a+n-1)}{b(b+1)(b+2)......(b+n-1)}$ converges or diverges ? where a>0 and b>a+1.
if the sum converges what is the sum i.e where it will converge? i need a concrete explanation
i found some inequalities $\frac{a(a+1)}{b(b+1)}$ <$\frac{a}{b+1}$ and $\frac{a(a+1)(a+2)}{b(b+1)(b+2)}$< $\frac{a}{b+2}$ .......and continuing $\frac{a(a+1)....(a+n-1)}{b(b+1)....(b+n-1)}$ < $\frac{a}{b+n-1}$
this inequalities can be useful for this problem
 A: Since Hamou has answered the question of convergence, I'll answer the question about the sum of the series. For $b - a > 1$, the sum of the series is $$-1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
Indeed, Abel's continuity theorem gives 
$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots (a + n-1)}{b(b + 1)\cdots (b + n - 1)} = \lim_{x\to 1^{-}} \sum_{n = 1}^\infty \frac{a(a+1)\cdots (a+n-1)}{b(b+1)\cdots (b+n-1)}x^n$$
For $\lvert x \rvert < 1$, 
$$\sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}x^n = \frac{\Gamma(b)}{\Gamma(a)}\sum_{n = 0}^\infty \frac{\Gamma(a+n)}{\Gamma(b+n)}x^n = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\sum_{n = 1}^\infty \frac{\Gamma(b-a)\Gamma(a+n)}{\Gamma(b+n)}x^n$$
The quotients $\Gamma(b-a)\Gamma(a+n)/\Gamma(b+n)$ are represented by Beta integrals
$$\int_0^1 (1 - t)^{b-a-1}t^{a+n-1}\, dt$$
Thus
$$\sum_{n = 0}^\infty \frac{\Gamma(b-a)\Gamma(a+n)}{\Gamma(b+n)}x^n = \int_0^1 \sum_{n = 0}^\infty (tx)^n(1-t)^{b-a-1}t^{a-1}\, dt = \int_0^1 (1 - tx)^{-1}(1-t)^{b-a-1}t^{a-1}\, dt$$
This gives the integral representation 
$$\sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}x^n = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\int_0^1 (1 - tx)^{-1}(1-t)^{b-a-1}t^{a-1}\, dt$$
Taking the limit as $x\to 1^{-}$ results in 
$$\frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)} \int_0^1 (1 - t)^{b-a-2}t^{a-1}\, dt = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\frac{\Gamma(b-a-1)\Gamma(a)}{\Gamma(b-1)} = \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
So 
$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)} = -1 + \sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)} = -1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
A: Let $u_n=\dfrac{a(a+1)\ldots(a+n-1)}{b(b+1)\ldots(b+n-1)}$
We have $\dfrac{u_{n+1}}{u_n}=1-\dfrac{b-a}{n}+o(\dfrac{1}{n})$
Hence by Raabe Duhamel  :
If $b-a>1$ the infinite sum converge
If $b-a<1$ the infinite sum diverge.
