# Gamma Function- Limit/ Product Representation-

• May 8th 2010, 05:07 PM
Anthonny
Gamma Function- Limit/ Product Representation-
Hello,

I am trying to figure out how the product representation of the Gamma function can be rewritten as this limit.

I've tried working on it for hours and I still cannot figure it out.

First I worked trying to change the product representation into the limit, and the other way around and still couldn't figure it out.

The limit representation is given by:

$\Gamma(z)=\lim_{n\rightarrow\infty}\frac{n!n^z}{z( z+1)(z+2)...(z+n)}$

The product representation is given by:

$\Gamma(z)=\frac{1}{z}\prod^\infty_{n=1}\frac{(1+\f rac{1}{n})^z}{1+\frac{z}{n}}$

I understand how the denominators are equal after some manipulation, but I can't quite understand the numerators.

Trying to manipulate the limit definition, this is what I've done:

$\Gamma(z)=\frac{1}{z}\lim_{n\rightarrow\infty}\fra c{n!n^z}{(z+1)(z+2)...(z+n)}$

I then just focused on changing the denominator of the limit into an infinite product.

$\frac{1}{z}\lim_{n\rightarrow\infty}\frac{1}{(z+1) (z+2)...(z+n)}=\frac{1}{z}\prod^\infty_{n=1}\frac{ 1}{z+n}$

$\frac{1}{z}\prod^\infty_{n=1}\frac{1}{z+n}\frac{\f rac{1}{n}}{\frac{1}{n}}=\frac{1}{z}\prod^\infty_{n =1}\frac{n^{-1}}{(1+\frac{z}{n})}$

So I see how the denominators resemble each other. However, I can't quite figure out the numerators.

I've done some more manipulations with the numerator which I won't show here out of tediousness, but the work didn't get me anywhere.

It would be greatly appreciated if an explanation was posted on how to convert the given infinite product definition into the limit definition, or the limit definition into the infinite product.

Thank you.
• May 8th 2010, 05:40 PM
Bruno J.
You need a good trick of Weierstrass! There's no way you will get from one to the other by just rearranging what's there in a trivial fashion.

In the limit representation, write $n^z$ as $e^{z\log n}$ and use the fact that $\log n = 1+\frac{1}{2}+\dots + \frac{1}{n} - \gamma + o(1)$ where $o(1)\rightarrow 0$ as $n\rightarrow \infty$.
• May 8th 2010, 05:43 PM
Bruno J.
I corrected a small mistake in my post (forgot gamma constant).

Edit : what I suggested above will yield Weierstrass' product representation, and not Euler's (which is the one you are looking for).
Still, probably a step in the good direction.
I'll give it some more thought!
• May 8th 2010, 06:01 PM
Anthonny
Quote:

Originally Posted by Bruno J.
I corrected a small mistake in my post (forgot gamma constant).

Edit : what I suggested above will yield Weierstrass' product representation, and not Euler's (which is the one you are looking for).
Still, probably a step in the good direction.
I'll give it some more thought!

Thanks!
I see what you mean on the faultiness of trying to rearrange the two representations. I didn't quite realize that what I was doing wasn't readily attainable through the methods I used.

I will take your suggestion in the Weierstrass' product representation. I've seen it before, but haven't quite worked at it.

• May 8th 2010, 06:43 PM
Bruno J.
Here's an idea. Can you show that $\lim_{n \rightarrow \infty} n^{-z}\prod_{k=1}^n\left(1+\frac{1}{k}\right)^z = 1$? It would follow from this; because, letting

$A_n = \frac{1}{z}\prod_{k=1}^n \frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}}$

$B_n = \frac{n!n^z}{z(z+1)\dots(z+n)}$

we just want to show that $\lim_{n \rightarrow \infty}A_n/B_n = 1$, which easily reduces to the above.
• May 8th 2010, 08:17 PM
Anthonny
Quote:

Originally Posted by Bruno J.
Here's an idea. Can you show that $\lim_{n \rightarrow \infty} n^{-z}\prod_{k=1}^n\left(1+\frac{1}{k}\right)^z = 1$? It would follow from this; because, letting

$A_n = \frac{1}{z}\prod_{k=1}^n \frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}}$

$B_n = \frac{n!n^z}{z(z+1)\dots(z+n)}$

we just want to show that $\lim_{n \rightarrow \infty}A_n/B_n = 1$, which easily reduces to the above.

Thank you!

I've tried what you suggested. I'll just post my work here:

$A_n=\lim_{n\rightarrow\infty}\frac{1}{z}\prod^n_{k =1}\frac{(1+\frac{1}{k})^z}{(1+\frac{z}{k})}$

$B_n=\lim_{n\rightarrow\infty}\frac{n!n^z}{z(z+1).. .(z+n)}$

$\frac{A_n}{B_n}=\lim_{n\rightarrow\infty}\frac{1}{ n^z}\prod^n_{k=1}\frac{(1+\frac{1}{k})^z}{(1+\frac {z}{k})}\cdot\frac{(z+1)(z+2)...(z+n)}{(1)(2)(3).. .(n-1)(n)}$

$\frac{A_n}{B_n}=\lim_{n\rightarrow\infty}\frac{1}{ n^z}\prod^n_{k=1}\frac{k(1+\frac{1}{k})^z}{(z+k)}\ cdot\prod^n_{k=1}\frac{z+k}{k}=\lim_{n \rightarrow \infty} n^{-z}\prod_{k=1}^n\left(1+\frac{1}{k}\right)^z$

Then evaluating the limit:

$\lim_{n \rightarrow \infty} n^{-z}\prod_{k=1}^n\left(1+\frac{1}{k}\right)^z=\lim_{ n \rightarrow \infty} n^{-z}(1+1)^z(1+\frac{1}{2})^z(1+\frac{1}{3})^z(1+\fra c{1}{4})^z...(1+\frac{1}{n})^z$

$=\lim_{n \rightarrow \infty} n^{-z}\left(2*\frac{3}{2}*\frac{4}{3}*\frac{5}{4}*...\ frac{n+1}{n}\right)^z$

$=\lim_{n \rightarrow \infty} n^{-z}(n+1)^z=\lim_{n \rightarrow \infty} \left(\frac{n+1}{n}\right)^z=\lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^z=1^z=1$

And so because the limit of the quotient of these two expressions are equal to 1, then these two expressions must be equal.

Everything seems to check out and work. Again thank you. It's greatly appreciated.
• May 8th 2010, 09:19 PM
Bruno J.
You're welcome! Good work.

Note also that we know both expressions have the same limit because we know one of them converges. Otherwise it wouldn't be enough for the limit of their ratio to be $1$.