Limit of infinite product

**Serg777** · February 22nd 2009, 09:14 AM

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.

**Danny** · February 22nd 2009, 09:52 AM

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

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

**Serg777** · February 22nd 2009, 09:58 AM

Yes.I am.

**Danny** · February 22nd 2009, 10:15 AM

Try this

$<br /> \lim_{n \to \infty} \frac{(n+1)^{\frac{1}{n}} (n+2)^{\frac{1}{n}} \cdots (2n-1)^{\frac{1}{n}} (2n)^{\frac{1}{n}} }{n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} \cdots \qquad \;\;n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} }$

then group separately. So you're really looking at

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

**Serg777** · February 22nd 2009, 10:20 AM

Thank you VERY MUCH

**Serg777** · February 22nd 2009, 10:39 AM

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

**Moo** · February 22nd 2009, 10:48 AM

Originally Posted by Serg777

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

$\lim_{n \to \infty} \frac{(n+1)^{\frac{1}{n}} (n+2)^{\frac{1}{n}} \cdots (2n-1)^{\frac{1}{n}} (2n)^{\frac{1}{n}} }{n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} \cdots \qquad \;\;n^{\frac{1}{n}} \qquad \;\;n^{\frac{1}{n}} }<br />$
$=\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{1/n} \cdots \left(\frac{2n-1}{n}\right)^{1/n} \left(\frac{2n}{n}\right)^{1/n}$

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

**Serg777** · February 22nd 2009, 11:00 AM

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?

**Serg777** · February 22nd 2009, 11:02 AM

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

**Krizalid** · February 22nd 2009, 11:02 AM

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

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

**Serg777** · February 22nd 2009, 11:08 AM

Thanks, fellows. You are great.

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