Evaluate $\lim_{n\to\infty}\frac {qn+1}{qn} \frac {qn+p+1}{qn+p} \cdots \frac {qn+np+1}{qn+np}$ where $p\in\mathbb{N}\setminus\{0,1\},q>0.$

Question

Evaluate $$\lim_{n\to\infty}\frac {qn+1}{qn} \frac {qn+p+1}{qn+p} \dotsm \frac {qn+np+1}{qn+np}$$ where $p\in\mathbb{N}\setminus\{0,1\},q>0.$

Any hints on how to approach this problem in the first place? The answer should be: $\left(\frac {p+q}q\right)^{1/p}$

I have just tried something and got pretty close to the result anyway:

Let $\lim_{n\to\infty} \frac {qn + 1}{qn}\frac {qn+p+1}{qn+p}\dotsm\frac {qn+np+1}{qn+np}=L$ taking the $\ln$ on both sides we get:

$$\lim_{n\to\infty} \sum_{k=0}^n\ln\left(1+\frac1{qn+kp}\right).$$

Here (I'm not sure if I'm wrong here) but if we use the remarkable limit on each of these $\ln$. $\lim_{x\to0}\frac{ln(1+x)}{x}=1$.

$$\lim_{n\to\infty}\sum_{k=0}^n\frac1n\frac1{q+ \frac knp}\xrightarrow[{n\rightarrow\infty}]{}\int_0^1\frac 1{q+xp}dx=\frac 1p\ln\frac {p+q}{q}=\ln L.$$

$\implies \boxed{L=\sqrt[p]{\frac{p+q}q}}.$

Answer 1

Gamma Function Approach $$ \begin{align} \lim_{n\to\infty}\prod_{k=0}^n\frac{qn+pk+1}{qn+pk} &=\lim_{n\to\infty}\prod_{k=0}^n\frac{\frac{qn+1}p+k}{\frac{qn}p+k}\tag1\\ &=\lim_{n\to\infty}\frac{\Gamma\!\left(\frac{qn+1}p+n+1\right)}{\Gamma\!\left(\frac{qn+1}p\right)}\frac{\Gamma\!\left(\frac{qn}p\right)}{\Gamma\!\left(\frac{qn}p+n+1\right)}\tag2\\ &=\lim_{n\to\infty}\frac{\left(\frac{p+q}pn+1\right)^{1/p}}{\left(\frac{q}pn\right)^{1/p}}\tag3\\[6pt] &=\left(\frac{p+q}q\right)^{1/p}\tag4 \end{align} $$ Explanation:
$(1)$: divide the numerators and denominators by $p$
$(2)$: write the product in terms of the Gamma function
$(3)$: Gautshi's Inequality gives that $\lim\limits_{n\to\infty}\frac{\Gamma(n+x)}{n^x\,\Gamma(n)}=1$
$(4)$: evaluate the limit

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Riemann Sum Approach $$ \begin{align} \lim_{n\to\infty}\log\left(\prod_{k=0}^n\frac{qn+pk+1}{qn+pk}\right) &=\lim_{n\to\infty}\sum_{k=0}^n\log\left(1+\frac1{qn+pk}\right)\tag5\\ &=\lim_{n\to\infty}\sum_{k=0}^n\left(\color{#090}{\frac1{q+p\frac kn}\frac1n}+\color{#C00}{O\left(\frac1{n^2}\right)}\right)\tag6\\ &=\color{#090}{\int_0^1\frac1{q+px}\,\mathrm{d}x}+\color{#C00}{0}\tag7\\[6pt] &=\frac1p\log\left(\frac{p+q}q\right)\tag8 \end{align} $$ Explanation:
$(5)$: log of a product is the sum of the logs
$(6)$: $\log(1+x)=x+O\!\left(x^2\right)$
$(7)$: green part is the Riemann Sum for the integral, the red part is $O\!\left(\frac1n\right)$
$(8)$: evaluate the integral

Answer 2

"if we use the remarkable limit on each of these. $\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$". This is a nice approach, but it isn't entirely rigorous. Fortunately we only need the squeeze/sandwich theorem to make this rigorous:

Either from the Taylor series, $\ln(1+x) = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{x^k}{k} = x - x^2/2 + x^3/3 -\dotsb$, or by other means, for $x>0$: $$x-\frac{x^2}{2} \leq \ln(1+x) \leq x$$

So we obtain bounds on the sum: $$\sum_{k=0}^n \frac{1}{qn+pk} - \frac12 \sum_{k=0}^n \left(\frac{1}{qn+pk}\right)^2 \leq \sum_{k=0}^n \ln\left(1 + \frac{1}{qn+pk}\right) \leq \sum_{k=0}^n \frac{1}{qn+pk} $$

As you pointed out, we may rewrite \begin{align} \sum_{k=0}^n \frac{1}{qn+pk} &= \sum_{k=0}^n \frac1n \frac{1}{q+pk/n} \to \int_0^1\frac{dx}{q+px} = \frac1p \ln\frac{p+q}{q}\\ n\sum_{k=0}^n \left(\frac{1}{qn+pk}\right)^2 &= \sum_{k=0}^n \frac{1}{n}\left(\frac{1}{q+pk/n}\right)^2 \to \int_0^1 \frac{dx}{(q+px)^2} =\frac{1}{q(p+q)} \end{align} so $$\lim_{n\to \infty}-\frac12 \sum_{k=0}^n \left(\frac{1}{qn+pk}\right)^2 = 0$$ Finally, by the squeeze theorem:

$$\lim_{n\to\infty} \sum_{k=0}^n \ln\left(1 + \frac{1}{qn+pk}\right) = \lim_{n\to\infty} \sum_{k=0}^n \frac{1}{qn+pk} = \frac1p \ln \frac{p+q}{q}$$

Answer 3

Hint: write your product as

$$\frac{\Gamma \left(\frac{n q}{p}+1\right) \Gamma \left(\frac{q n}{p}+n+\frac{1}{p}+1\right)}{\Gamma \left(\frac{q n}{p}+n+1\right) \Gamma \left(\frac{n q}{p}+\frac{1}{p}+1\right)}$$

where $\Gamma(s)$ is the Gamma function, and use the asymptotic behaviour

$$\frac{\Gamma(n+\alpha)}{\Gamma(n)}\sim n^\alpha\qquad(n\to\infty).$$

Answer 4

Here is another approach, which does not require the gamma function nor integrals.

We have

$$\prod_{a=1}^p\prod_{k=1}^n\frac{qn+pk+a}{qn+pk+a-1} =\frac{n p}{n q+p}+1\to \frac{p+q}q\qquad(n\to\infty)$$

since the product is telescoping. Hence, to derive the desired result it suffices to show that the limit

$$f(a)=\lim_{n\to\infty}\prod_{k=1}^n\frac{qn+pk+a+1}{qn+pk+a} =\lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{1}{qn+pk+a}\right)$$

is independent of $a$. We only need the elementary inequalities

$$\frac1{1+\frac1x}\leq\log(1+x)\leq x\quad(x\geq0)$$

which can be derived showing that the functions $\log(1+x)-\frac1{1+\frac1x}$ and $x-\log(1+x)$ are increasing for $x\geq0$. Thus, if we let $y=qn+pk+a$,

$$ \begin{align} 0\leq\log\left(\frac{f(a)}{f(a+1)}\right) &=\lim_n\sum_{k=1}^n\left[\log\left(1+\frac{1}{y}\right) -\log\left(1+\frac{1}{y+1}\right)\right] \\&\leq \lim_n\sum_{k=1}^n\left[\frac1y-\frac1{y+2}\right] \\&=2\lim_n\sum_{k=1}^n\frac1{y(y+2)} \\&\leq2\lim_n\sum_{k=1}^n\frac1{y^2} \end{align}$$

and since $y\geq qn$, it follows that

$$0\leq\log\left(\frac{f(a)}{f(a+1)}\right) \leq 2\lim_n\frac{1}{q^2n^2}\sum_{k=1}^n1 =\frac{2}{q^2n}\to0\qquad(n\to\infty).$$

This means that $f(a+1)=f(a)$ for all $a$, hence $f(1)=f(2)=\cdots=f(p)$, as wanted.