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

Thank a lot !

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up vote 4 down vote accepted

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 $$ \lim\limits_{n\to\infty}R_n=\left(\frac{a+1}a\right)^b=\left(\frac{q+p}q\right)^{1/p}. $$

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W|P is your friend. –  Did Jul 18 '12 at 12:15
    
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
    
@joriki: The Edit clarifies this, I think. –  Did Jul 18 '12 at 13:30
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