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 $\lim\limits\_{n\to\infty} p\_n = 0$.
Let $L=\lim\limits\_{n\to\infty}p\_n.$ If $b\ge a+1$ the following inequalities hold: $$\require\cancel 0<p\_n\le \frac{a\cancel{(a+1)\cdots(a+n)}}{\cancel{(a+1)(a+2)\cdots(a+n)}(a+1+n)}=\frac{a}{a+1+n}\to0,$$ and thus $L=0$ by the squeeze theorem.

For the case $b<a+1$, note that since for any positive $m$ $$\frac{a}{b}<\frac{a+m}{b+m},$$ we have $$L\le \lim\_{n\to\infty}\left(\frac{a+n}{b+n}\right)^n=\lim\_{n\to\infty}\left(1+\frac{a-b}{b+n}\right)^n=e^{a-b}<1.$$ Now suppose $L>0$. Then, $$\lim\_{n\to\infty}p\_n^{p\_n}=L^L<1\tag{$\star$}.$$ But $$\lim\_{n\to\infty}p\_n^{p\_n}\ge\lim\_{n\to\infty} \left(\frac{a\cancel{(a+1)\cdots(a+n)}}{\cancel{(a+1)(a+2)\cdots(a+n)}(a+1+n)}\right)^{\frac{a\cancel{(a+1)\cdots(a+n)}}{\cancel{(a+1)(a+2)\cdots(a+n)}(a+1+n)}} \\ \lim\_{n\to\infty}p\_n^{p\_n}\ge\lim\_{n\to\infty} \left(\frac{a}{a+1+n}\right)^{\frac{a}{a+1+n}}=1,$$ which contradicts $(\star)$. Therefore, $L=0$.