Question: I am solving another problem and it boils down to solving this problem:

We have that $b > a > 0$ are two arbitrary given real numbers.

Let $p\_n = \dfrac{a(a+1)...(a+n)}{b(b+1)...(b+n)}$

Prove that 

$\displaystyle\lim\_{n\to\infty} p\_n = 0$.

Is this true at all?  

How do we prove it?

Answer: 
To prove that $\lim\limits\_{n\to \infty} p\_n = 0$, we can, since $p\_n > 0$ for all $n$, show that $\lim\limits\_{n\to\infty} \frac{1}{p\_n} = +\infty$.

Rewriting in a convenient manner, we obtain

\begin{align}
\frac{1}{p\_n} &= \prod\_{k=0}^n \frac{b+k}{a+k}\\
&= \prod\_{k=0}^n \biggl(1 + \frac{b-a}{a+k}\biggr).
\end{align}

Now we use that for $x\_k \geqslant 0$ we have

$$\prod\_{k=0}^n (1 + x\_k) \geqslant 1 + \sum\_{k=0}^n x\_k.\tag{$\ast$}$$

The inequality $(\ast)$ is easily proved by induction. Using $(\ast)$, we deduce

$$\frac{1}{p\_n} \geqslant 1 + \sum\_{k=0}^n \frac{b-a}{a+k}.\tag{1}$$

Comparison with the harmonic series shows that

$$\lim\_{n\to\infty} \sum\_{k=0}^n \frac{b-a}{a+k} = +\infty,$$

and hence $\lim\limits\_{n\to\infty} p\_n = 0$.
One may recall that, as $n \to +\infty$, by the [generalized Stirling formula](https://math.stackexchange.com/questions/947219/equivalent-of-xx1x2-cdotsxn/952122#952122), we have
$$
a(a+1)(a+2)\cdots(a+n) \sim \frac{n^{n+a}e^{-n}\sqrt{2\pi n} }{\Gamma(a)} 
$$
giving $$
 p\_n = \frac{a(a+1)...(a+n)}{b(b+1)...(b+n)} \sim \frac{n^{n+a}e^{-n}\sqrt{2\pi n} }{\Gamma(a)}\cdot \frac{\Gamma(b) }{n^{n+b}e^{-n}\sqrt{2\pi n}}
$$ that is, as $n \to +\infty$: $$
 p\_n = \frac{a(a+1)...(a+n)}{b(b+1)...(b+n)} \sim \frac{\Gamma(b) }{\Gamma(a)}n^{a-b} \to 0,
$$ since $0<a<b$.