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https://proofwiki.org/wiki/Zeroes_of_Gamma_Function

Note that $\Gamma(z)= \frac{1}{z} \prod_{n=1}^\infty \frac{(1+1/n)^z}{1+z/n}$. Suppose $\Gamma(z)=0$ for some $z$ in the domain.

The proof in the link says that since each term of the product tends to $1$, the infinite product cannot be zero. However, I think this is a false proof. As far as I know, there are examples such that the terms tend to $1$ but the infinite product might be zero. Am I right?

Moreover, is there a way to prove that the Gamma function does not have a zero instead of using the equation $\Gamma(z)\Gamma(1-z)=\frac{1}{\sin \pi z}$?

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  • The proof in the link is not correct (or at least not complete), since $a_n\rightarrow 1$ and $a_n\neq 0$ doesn't imply $\prod_{n=1}^\infty a_n\neq 0$, see for example $a_n=\frac{n}{n+1}$. So yes, you are right. – Wojowu Mar 26 '16 at 11:00

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The proof given is indeed flawed. However, we can fix the argument with a bit of work.

For a given $z$, $$ \begin{align} \log\left(\frac{\left(1+\frac1k\right)^z}{1+\frac zk}\right) &=z\log\left(1+\frac1k\right)-\log\left(1+\frac zk\right)\\ &=z\left(\frac1k+O\left(\frac1{k^2}\right)\right)-\left(\frac zk+O\left(\frac1{k^2}\right)\right)\\ &=O\left(\frac1{k^2}\right) \end{align} $$ Therefore, since $\sum\limits_{k=1}^\infty\frac1{k^2}$ converges, $$ \log\left(\prod_{k=1}^\infty\frac{\left(1+\frac1k\right)^z}{1+\frac zk}\right) =\sum_{k=1}^\infty\log\left(\frac{\left(1+\frac1k\right)^z}{1+\frac zk}\right) $$ converges absolutely.

This means that the product cannot be $0$ unless one of the terms is equal to $0$.

We can verify the formula given for $\Gamma(z)$. $$ \begin{align} \prod_{k=1}^n\frac{\color{#C00000}{\left(1+\frac1k\right)^z}}{\color{#00A000}{1+\frac zk}} &=\color{#C00000}{(n+1)^z}\color{#00A000}{\frac{\Gamma(1+z)\,\Gamma(n+1)}{\Gamma(n+z+1)}}\\ &=\frac{(n+1)^z}{n+z}\frac{\Gamma(n+1)}{\Gamma(n+z)}\Gamma(1+z) \end{align} $$ Using Gautschi's Inequality, we get $$ \frac{(n+1)^z}{n+z}n^{1-z} \le\frac{(n+1)^z}{n+z}\frac{\Gamma(n+1)}{\Gamma(n+z)} \le\frac{(n+1)^z}{n+z}(n+1)^{1-z} $$ Therefore, by the Squeeze Theorem, $$ \lim_{n\to\infty}\frac{(n+1)^z}{n+z}\frac{\Gamma(n+1)}{\Gamma(n+z)}=1 $$ Thus, $$ \bbox[5px,border:2px solid #C0A000]{\frac1z\prod_{k=1}^\infty\frac{\left(1+\frac1k\right)^z}{1+\frac zk}=\Gamma(z)} $$

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