hard time coming up with an idea to find a limit 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?
 A: 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 $\lim\limits_{n\to\infty} p_n = 0$.
A: One may recall that, as $n \to +\infty$, by the generalized Stirling formula, 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 $0<a<b$.
A: 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, $L=0$.
