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In the Portuguese book Análise Matemática (Mathematical Analysis) by C. Sarrico, it is proved there exist

$$\displaystyle\lim_{n\rightarrow +\infty}f_{n}(x)=\lim_{n\rightarrow +\infty }e^{\log f_{n}(x)},\qquad(1)$$

where

$$f_{n}(x)=\dfrac{n!n^{x}}{x(x+1)(x+2)\cdots (x+n)}.\qquad(2)$$

And it is stated there that

$$\displaystyle\lim_{n\rightarrow +\infty }f_{n}(x)=\Gamma (x)=\int_{0}^{+\infty}t^{x-1}e^{-t}dt\qquad x>0.\qquad(3)$$

The author writes that a proof can be found in "Principles of Mathematical Analysis" by W. Rudin. Since I don't have it, I ask the following:

Question: How is the sketch of such a proof or other proofs of the same result?

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2 Answers

up vote 4 down vote accepted

I don't have a copy of Rudin, but can give you a proof of this. There's various ways of playing about with the integral form to get what you want, although I'm not sure what the cleanest method is. One way is to use the fact that $1_{\{t\le n\}}(1-t/n)^n\to e^{-t}$ to write $\Gamma(x)$ as $$ \begin{align} \Gamma(x)&=\lim_{n\to\infty}\int_0^nt^{x-1}(1-t/n)^{n}\,dt\\ &=\lim_{n\to\infty}n^x\int_0^1s^{x-1}(1-s)^n\,ds. \end{align} $$ To prove that the first integral commutes with the limit, you could use the dominated convergence theorem. The second integral is just using the substitution $t=ns$. It needs to be shown that this integral is equal to your function $f_n(x)$. In fact, I recognize both the function $f_n$ and the integral as beta functions. $$ \int_0^1s^{x-1}(1-s)^n\,ds=B(x,n+1) $$ To find the explicit form for this, you can either use $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$ (a proof is given on the Wikipedia page) or repeatedly apply the identity $B(x,n+1)=B(x+1,n)n/x$ (which follows from integration by parts) and $B(x,1)=1/x$.

Note: You can also apply a very similar argument to the one above using the limit $(1+t/n)^{-n}\to e^{-t}$, which I did in my first version of this answer. Using $(1-t/n)^n\to e^{-t}$ seems cleaner though, so I edited the answer accordingly.

An alternative method is to write $$ \begin{align} \Gamma(x) &= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{\Gamma(n+1)\Gamma(x)}{\Gamma(x+n+1)}\\ &= \frac{\Gamma(x+n+1)}{\Gamma(n+1)}\frac{n!}{x(x+1)\cdots(x+n)}. \end{align} $$ Then, the limit you need would follow as long as it can be shown that $\Gamma(x+n+1)/\Gamma(n+1)$ approaches $n^x$ asymptotically as $n\to\infty$. $$ \Gamma(n+x+1)=\int_0^\infty t^x t^ne^{-t}\,dt $$ Then, the required expression depends on showing that, in the limit $n\to\infty$, to leading order, the integral only contributes for values of $t/n$ close to 1 (you can see that $t^ne^{-t}$ has its maximum at $t=n$).

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+1 for your excellent explanation! –  Américo Tavares Aug 26 '10 at 14:11
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First, it should be noted that the integral representation is valid only for x with positive real part.

Having gotten that disclaimer away, one sketch of a proof goes as follows: the limit representation can be shown to be equivalent to the Weierstrass infinite product:

$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{j=1}^\infty{\left(1+\frac{z}{j}\right)\exp\left(-\frac{z}{j}\right)}$

($\gamma$ here is the Euler-Mascheroni constant; the representation $\lim_{k\to\infty}{\sum_{j=1}^k{\frac1{j}}-\ln(k)}$ is useful here).

You now show that the infinite product/limit is the same as the integral $n^z \int_0^1 (1-u)^n u^{z-1} \mathrm{d}u$, and then find a substitution for which Watson's lemma can be applied.

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The sketch is from Temme's "Special Functions" book, if you're stuck, I'll reply with the "magic substitution" needed for that integral representation. –  J. M. Aug 26 '10 at 13:57
    
Another hint: try integrating by parts the integral I gave. –  J. M. Aug 26 '10 at 14:03
    
@J. Mangaldan: "the integral representation is valid only for x with positive real part" -- I have only thought on $x$ as a real. I would include in my question that $x>0$, which I've forgotten. –  Américo Tavares Aug 26 '10 at 14:03
    
That's okay too, I'm just noting the general region of validity for the integral representation. –  J. M. Aug 26 '10 at 14:09
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