Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Thank a lot !

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.