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Let $$\Gamma^{(n)}(x) \quad(n \in \mathbb{N}, x \in \mathbb{R})$$ be the notation for the Gamma Function $\Gamma(x)$ used $n$ times on itself.

(Example: $\Gamma^{(3)}(x) = \Gamma(\Gamma(\Gamma(x)))$

Why is $\lim_{n \to \infty} \Gamma^{(n)}(x) = 1$ for $x \in [1,2]$?

It think it could have something to do with the fact that $0 < x! \leq 1$ for $x \in [0,1]$ or with the definition that $0! = 1! = 1$, but I can't prove it. Can someone help me?

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    $\begingroup$ Just an idea, not a full answer: On $[1,2]$ we have that $\vert\Gamma'(x)\vert<0.6$ as can be seen when graphing it: wolframalpha.com/… This means that $\Gamma$ is a contraction on the complete metric space $[1,2]$. According to the Banach fixed point theorem, every sequence of the type specified in your question will then converge to the same, unique fixed point of $\Gamma$. Since $\Gamma(1)=1$, this fixed point is $1$. I suspect the proof of $\vert\Gamma'(x)\vert<0.6$ is the hard part, though. $\endgroup$ Commented Jul 20, 2022 at 12:40

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As was mentioned by Vercassivelaunos in comments, it's sufficient to prove that $\left|\Gamma'(x)\right| \leq c < 1$ for $x\in[1, 2]$. It's known that $$ \Gamma^{(n)}(x) = \int_0^{\infty} t^{x-1} e^{-t} \left(\ln t\right)^n \mathrm{d}t, $$ so $\Gamma '' (x) > 0$ for all $x > 0$ and hence $\Gamma'(x)$ is increasing for $x > 0$. Let's find $\Gamma'(1)$ and $\Gamma'(2)$. From the properties of the digamma function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$, for $n\in\mathbb{N}$ we have $$\Gamma'(n+1) = \Gamma(n+1) \cdot \left(\sum_{k=1}^{n} \frac{1}{k} - \gamma\right) = n! \cdot \left(\sum_{k=1}^{n} \frac{1}{k} - \gamma\right),$$ where $\gamma$ is the Euler–Mascheroni constant. So, $\Gamma'(1) = -\gamma$ and $\Gamma'(2) = 1 - \gamma$.

As $\Gamma'(x)$ is increasing and $\gamma \approx 0.577$, we have $-\gamma \leq \Gamma'(x) \leq 1 - \gamma$ for $x\in[1, 2]$, which leads to $\left|\Gamma'(x)\right| \leq \gamma < 1$ for $x\in [1, 2]$. So, according to the Banach fixed point theorem, $\Gamma(x)$ is a contraction on $[1, 2]$ and has a unique fixed point $x^* \in [1, 2]$ such that $x^* = \Gamma(x^*)$, which can be found by iterating $x_{n+1} = \Gamma(x_n)$ for any $x\in[1, 2]$.

EDIT: just as an interesting fact, it turns out that there are many other regions in the complex plane such that $z_{n+1} = \Gamma(z_n)$ converges to 1: plot for $\mathrm{Re} \; z, \; \mathrm{Im} \; z \in [-5, 5]$ (although it says "absolute value", the limit equals real number 1): enter image description here Here is also zoomed out version for $\mathrm{Re} \; z, \mathrm{Im} \; z \in [-20, 20]$, some pattern can be seen for $\mathrm{Re} \; z > 0$.

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  • $\begingroup$ I've just found out that the limit also goes to 1 for $x \in [1,3)$. The proof should stay almost the same, right? $\endgroup$ Commented Jul 21, 2022 at 12:01
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    $\begingroup$ @BlackSheep73 no, because $\Gamma'(3)$ is already greater than 1. But it will work, for example, in case of $x\in[1, 2.5]$, because $\psi(2.5) = -\gamma - 2\ln 2 + \frac{8}{3} \approx 0.703$ and $\Gamma(2.5) = \frac{3}{4}\sqrt{\pi} \approx 1.329$, hence $\Gamma'(2.5) \approx 0.934$. However, derivative condition is only a simpler case of Banach theorem, which is again provides only a sufficient condition. So, there may be other ways to prove for $x \in [2.5, 3)$. Also, take attention that Banach theorem works only for complete spaces and $[2.5, 3)$ is not complete. $\endgroup$ Commented Jul 21, 2022 at 13:52
  • $\begingroup$ @BlackSheep73 turns out that there are more points in the complex plane with this property, than one can probably think of. I've added some pictures to the answer. $\endgroup$ Commented Jul 21, 2022 at 16:38

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