Limit when n-> infinity

**m3th0dman** · April 15th 2009, 04:05 AM

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

$p\in N-\begin{Bmatrix}0, & 1 \end{Bmatrix} ,q\in N^\ast$

**Opalg** · April 15th 2009, 08:10 AM

Originally Posted by m3th0dman

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

$p\in N-\begin{Bmatrix}0, & 1 \end{Bmatrix} ,q\in N^\ast$

I tend to be mistrustful of "big O" arguments, but I think this one is okay.

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

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

Taking exponentials, $\lim_{n\to\infty}\prod_{k=0}^n \frac{qn+kp+1}{qn+kp} = \bigl(1+\tfrac pq\bigr)^{1/p}$ .

**m3th0dman** · April 15th 2009, 12:22 PM

Originally Posted by Opalg

I tend to be mistrustful of "big O" arguments, but I think this one is okay.

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

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

Taking exponentials, $\lim_{n\to\infty}\prod_{k=0}^n \frac{qn+kp+1}{qn+kp} = \bigl(1+\tfrac pq\bigr)^{1/p}$ .

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.

**Opalg** · April 16th 2009, 02:15 AM

Originally Posted by m3th0dman

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 simplify things.
In this is case it depends on 1/n^2, so I suppose it goes to 0 when n-> infinity.

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 $\tfrac12x^2$ . Put $x= \frac1{qn+kp}$ to see that the difference between $\ln\Bigl(1 + \frac1{qn+kp}\Bigr)$ and $\frac1{qn+kp}$ is less than $\frac1{2q^2n^2}$ . Adding n+1 such terms together, you find that the difference between $\sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr)$ and $\sum_{k=0}^n\frac1{qn+kp}$ is less than $\frac{n+1}{2q^2n^2}$ . This difference goes to 0 as n goes to infinity.

**m3th0dman** · April 16th 2009, 04:38 AM

Originally Posted by Opalg

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 $\tfrac12x^2$ . Put $x= \frac1{qn+kp}$ to see that the difference between $\ln\Bigl(1 + \frac1{qn+kp}\Bigr)$ and $\frac1{qn+kp}$ is less than $\frac1{2q^2n^2}$ . Adding n+1 such terms together, you find that the difference between $\sum_{k=0}^n\ln\Bigl(1 + \frac1{qn+kp}\Bigr)$ and $\sum_{k=0}^n\frac1{qn+kp}$ is less than $\frac{n+1}{2q^2n^2}$ . This difference goes to 0 as n goes to infinity.

I suppose that Taylor's formula is applied this way:
$ 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:
$ ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}+... $
and
$ \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.

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