Is there an elementary way to prove that the limit $$\lim_{n\rightarrow \infty}\frac{n!(n+1)^z}{(z+1)(z+2)\cdots (z+n)}$$ exists for all complex numbers $z$ other than the negative integers? I know this can be proved by standard methods of complex analysis, but I would like to know if there is a simple proof that would be accessible to Calculus students. More generally, I would be interested in an elementary development of the Gamma function as a function defined for all complex numbers other than the negative integers.
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2 Let $z$ be a complex number which is not a negative integer. Let $u_n=\frac{n!e^{z\ln(n+1)}}{(z+1) \ldots (z+n)}$. Then $\frac{u_{n+1}}{u_n}=\frac{(n+1)(n+2)^z}{(z+n+1)(n+1)^z}=\frac{n+1}{z+n+1}\left(1+\frac{1}{n+1}\right)^z=\left(1+\frac{z}{n+1}+O(1/n^2)\right)\left(1-\frac{z}{n+1}\right)=1+O(n^{-2})$. From this, it follows that the product of the $u_{n+1}/u_n$ convergea absolutely, so $u_n$ converges to a nonzero limit.
- $\begingroup$ Yes, that's what I was looking for, thanks... is there an easy way to see that the same limit is $z!$ if $z$ is a positive integer? $\endgroup$Math101– Math1012021-07-26 00:30:04 +00:00Commented Jul 26, 2021 at 0:30
- 1$\begingroup$ For $z=m \geq 0$, $u_n=\binom{n+m}{m}^{-1} \times (n+1)^m$, so that $u_n \sim (n^m/m!)^{-1} \times n^m=m!$. $\endgroup$Aphelli– Aphelli2021-07-26 06:57:59 +00:00Commented Jul 26, 2021 at 6:57