# Thread: Limit of infinite product

1. ## Limit of infinite product

Hello. I need help in calculating the limit.
Lim((sqrt((n+1)*(n+2)*(n+3)*...*(2n-1)*2*n) of n)/n), n=infinity.
Surely, it exists.
I can't understand how I shall express the root of the product in the nominator or should I. I can understand that there must be k*n in the nominator, but I don't know how to find this k. simple product of n is not suitable.
Great thanks for everybody who will try to help me.

2. Originally Posted by Serg777
Hello. I need help in calculating the limit.
Lim((sqrt((n+1)*(n+2)*(n+3)*...*(2n-1)*2*n) of n)/n), n=infinity.
Surely, it exists.
I can't understand how I shall express the root of the product in the nominator or should I. I can understand that there must be k*n in the nominator, but I don't know how to find this k. simple product of n is not suitable.
Great thanks for everybody who will try to help me.
Just for clarifaction, are you trying to find

$\displaystyle \lim_{n \to \infty} \frac{\left((n+1)(n+2) \cdots (2n-1)2n\right)^{\frac{1}{n}}}{n}$

3. Yes.I am.

4. Try this

$\displaystyle \lim_{n \to \infty} \frac{(n+1)^{\frac{1}{n}} (n+2)^{\frac{1}{n}} \cdots (2n-1)^{\frac{1}{n}} (2n)^{\frac{1}{n}} }{n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} \cdots \qquad \;\;n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} }$

then group separately. So you're really looking at

$\displaystyle \lim_{n \to \infty} \left( 1 + \frac{i}{n}\right)^{\frac{1}{n}}\;\;\; i = 1 \cdots n$

5. Thank you VERY MUCH

6. The last limit is the limit of the sum. Correct? Can you explain in full how to group?

7. Originally Posted by Serg777
The last limit is the limit of the sum. Correct? Can you explain in full how to group?
$\displaystyle \lim_{n \to \infty} \frac{(n+1)^{\frac{1}{n}} (n+2)^{\frac{1}{n}} \cdots (2n-1)^{\frac{1}{n}} (2n)^{\frac{1}{n}} }{n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} \cdots \qquad \;\;n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} }$
$\displaystyle =\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{1/n} \cdots \left(\frac{2n-1}{n}\right)^{1/n} \left(\frac{2n}{n}\right)^{1/n}$

In fact, it's the limit of $\displaystyle \lim_{n \to \infty} {\color{red}\prod_{i=1}^n} \left(1+\frac in\right)^{1/n}$, the limit of the product

8. I thought it was smth more complicated about grouping. Is it possible to reduce to sum? Can we omit everything except numbers and 1/(n+1)? By grouping (n+1)*(1+2/(1+n))^(1/n).....
Doing this we get (1+1/(n+1))*(1+2/(n+1).....). Of course, root of n.

What do you think?

9. Because I have no idea how to find the limit of product.

10. Following up on Moo's idea, we have that the product equals $\displaystyle \exp \left\{ \frac{1}{n}\sum\limits_{i=1}^{n}{\ln \left( 1+\frac{i}{n} \right)} \right\}\to \exp \left( \int_{1}^{2}{\ln x\,dx} \right)$ as $\displaystyle n\to\infty.$

There're a few ways of solving this last integral, the standar one, it's to apply integration by parts.

11. Thanks, fellows. You are great.