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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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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](http://en.wikipedia.org/wiki/Stirling%27s_formula#Stirling.27s_formula_for_the_Gamma_function) is your friend.2012-07-18
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    I think you applied it with $x=an+n$, not $an+n+1$? (In any case +1 :-)2012-07-18
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    @joriki: If my expression of $R_n$ is correct, one uses $x=an+n+1$. What am I missing? (In any case, thanks.)2012-07-18
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    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 $(a+1)^b$?2012-07-18
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    @joriki: The Edit clarifies this, I think.2012-07-18
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    It does, thanks.2012-07-18
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    Sure. The computations become messier but this can be done.2012-07-18
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    Can you help me , please ?2012-07-18