$\displaystyle \lim_{n->\infty}\frac{qn+1}{qn}\frac{qn+p+1}{qn+p}\frac{qn +2p+1}{qn+2p}...\frac{qn+np+1}{qn+np}$

$\displaystyle p\in N-\begin{Bmatrix}0,

& 1

\end{Bmatrix}

,q\in N^\ast $

Printable View

- Apr 15th 2009, 04:05 AMm3th0dmanLimit when n-> infinity
$\displaystyle \lim_{n->\infty}\frac{qn+1}{qn}\frac{qn+p+1}{qn+p}\frac{qn +2p+1}{qn+2p}...\frac{qn+np+1}{qn+np}$

$\displaystyle p\in N-\begin{Bmatrix}0,

& 1

\end{Bmatrix}

,q\in N^\ast $ - Apr 15th 2009, 08:10 AMOpalg
I tend to be mistrustful of "big O" arguments, but I think this one is okay.

Notice that $\displaystyle \frac{qn+kp+1}{qn+kp} = 1 + \frac1{qn+kp}$, and $\displaystyle \ln\Bigl(1 + \frac1{qn+kp}\Bigr) = \frac1{qn+kp} + O(n^{-2})$. Therefore

$\displaystyle \begin{aligned}\sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr) &= \sum_{k=0}^n\frac1{qn+kp} + O(n^{-1})\\ &= \frac1n\sum_{k=0}^n\frac1{q+\frac knp} + O(n^{-1})\\ &\to \int_0^1\frac1{q+xp}\,dx\qquad\text{(limit of Riemann sum)}\\ &= \tfrac1p\ln\bigl(1+\tfrac pq\bigr).\end{aligned}$

Taking exponentials, $\displaystyle \lim_{n\to\infty}\prod_{k=0}^n \frac{qn+kp+1}{qn+kp} = \bigl(1+\tfrac pq\bigr)^{1/p}$. - Apr 15th 2009, 12:22 PMm3th0dman
Thank you very much. This is the correct answer (written at the end of the Book). Unfortunately I am not familiar with Big O notation, so I will have to chew on it a little.

I would be grateful if you could explain it without "Big O" notation, or if easier you could explain the "Big O" notation.

I understand that is a function that is not important in finding the answer, written to simplfy things.

In this is case it depends on 1/n^2, so I supouse it goes to 0 when n-> infinity. - Apr 16th 2009, 02:15 AMOpalg
The big O notation is just a succinct way of estimating orders of magnitude. For this problem, you can make the estimate explicit by using Taylor's theorem with remainder for the function ln(1+x). This tells you that, for 0<x<1, the difference between ln(1+x) and x is less than $\displaystyle \tfrac12x^2$. Put $\displaystyle x= \frac1{qn+kp}$ to see that the difference between $\displaystyle \ln\Bigl(1 + \frac1{qn+kp}\Bigr)$ and $\displaystyle \frac1{qn+kp}$ is less than $\displaystyle \frac1{2q^2n^2}$. Adding n+1 such terms together, you find that the difference between $\displaystyle \sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr)$ and $\displaystyle \sum_{k=0}^n\frac1{qn+kp}$ is less than $\displaystyle \frac{n+1}{2q^2n^2}$. This difference goes to 0 as n goes to infinity.

- Apr 16th 2009, 04:38 AMm3th0dman
I suppose that Taylor's formula is applied this way:

$\displaystyle

ln(1+x)=ln(a+1)+\frac{x-a}{1+a}-\frac{(x-a)^2}{2!(1+a)^2}+\frac{(x-a)^3}{3!(1+a)^4}-...

$

And we apply it for a=0:

$\displaystyle

ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}+...

$

and

$\displaystyle

\frac{x^2}{2} \geq -\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}+...

$

for x between [0,1].

I haven't studied Taylor's Theorem (or formula) at school, so I don't know if I had applied correctly.

Assuming what have I written is correct, I understood everything. Again I thank you very much.