# What is the limit for the following sequence of products?

$\lim\limits_{n \to \infty}$ $\displaystyle\frac{q \cdot n +1}{q \cdot n} \cdot \frac{q \cdot n +p+1}{q \cdot n +p} \cdot \ldots \cdot \frac{q \cdot n +n \cdot p +1}{q \cdot n + n \cdot p}$ , for $q > 0, p \geq 2$ .

Thank a lot !

-

Introducing the parameters $a=q/p$ and $b=1/p$, the $n$th ratio is $$R_n=\prod_{k=0}^n\frac{an+b+k}{an+k}=\frac{\Gamma(an+b+n+1)\cdot\Gamma(an)}{\Gamma(an+b)\cdot\Gamma(an+n+1)}.$$ One knows that $\Gamma(x+b)\sim x^b\cdot\Gamma(x)$ when $x\to+\infty$. Applying this twice, one gets $$\frac{\Gamma(an+b+n+1)}{\Gamma(an+n+1)}\sim (an+n+1)^b\sim (a+1)^bn^b, \qquad \frac{\Gamma(an+b)}{\Gamma(an)}\sim a^bn^b,$$ hence $$\lim\limits_{n\to\infty}R_n=\left(\frac{a+1}a\right)^b=\left(\frac{q+p}q\right)^{1/p}.$$
I think you applied it with $x=an+n$, not $an+n+1$? (In any case +1 :-) – joriki Jul 18 '12 at 13:09
@joriki: If my expression of $R_n$ is correct, one uses $x=an+n+1$. What am I missing? (In any case, thanks.) – Did Jul 18 '12 at 13:19
Using $x=an+n+1$, I get $(n(a+1)+1)^b$ instead of $(n(a+1))^{b+1}/(n(a+1))^1=(n(a+1))^b$, which I think is what's needed to cancel $n^b$ and be left with $(a+1)^b$? – joriki Jul 18 '12 at 13:22