How to evaluate $\lim\limits_{x\to \infty }\Gamma(x+2)^{\frac{1}{x+1}}-\Gamma(x+1)^{\frac{1}{x} }$

Question

This problem is in my problem book:

$$L:=\lim\limits_{x\to \infty }\Gamma(x+2)^{\frac{1}{x+1}}-\Gamma(x+1)^{\frac{1}{x} }$$

Whoever the book didn't provide a solution.

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My attempt:

I will use $x!=\Gamma(x+1)$

$$=\lim\limits_{x\to \infty }(x+1)!^{\frac 1{x+1}}-(x!)^{1/x}$$

$$=\lim\limits_{x\to \infty }x!^{1/x}\left(\frac{(x+1)^\frac{1}{x+1}-(x!)^{\frac{1}{x(x+1)}}}{(x!)^{\frac{1}{x(x+1)}}}\right)$$ $$=\lim\limits_{x\to \infty }\frac{x!^{1/x}}{x} \times x\left({(x+1)^\frac{1}{x+1}-(x!)^{\frac{1}{x(x+1)}}}\right)$$ $$\lim\limits_{x\to \infty }\frac{x^{1/x}}{x} =\frac 1 e$$

$$J:=x\left({(x+1)^\frac{1}{x+1}-(x!)^{\frac{1}{x(x+1)}}}\right)$$ $L=\frac J e$

Assume $x^{\frac1 {1+x}} b(x)=(x!)^{\frac{1}{x(x+1)}}$, $b(x) \to 1 $ as $x\to \infty$ $$J=\lim\limits_{x\to \infty }xb(x)\left({(x+1)^\frac{1}{x+1}-(x)^{\frac{1}{(x+1)}}}\right)+\lim\limits_{x\to \infty }x(x+1)^\frac{1}{x+1}\left(1-b(x)\right) $$

$$\lim\limits_{x\to \infty }x\left({(x+1)^\frac{1}{x+1}-(x)^{\frac{1}{(x+1)}}}\right)=0 \tag 1$$

$$J=\lim\limits_{x\to \infty} x \left(1-\left(\frac{x!}{x^x}\right)^{\frac{1}{x(x+1)} }\right) \tag 2$$

I couldn't prove that (1) would be zero and I couldn't evaluate (2).

Answer 1

Let $f(x) = \frac{1}{x}\log\Gamma(x+1)$ and note that

$$ xf'(x) = \psi(x+1) - \frac{\log x!}{x} = 1 + \mathcal{O}\left(\frac{\log x}{x}\right). $$

as $x \to \infty$. Here, $\psi$ is the digamma function and the last step follows from the Stirling's approximation and the asymptotic behavior of $\psi$ for large argument. From this and the Stirling's approximation again, we get

$$ [(x!)^{1/x}]' = \frac{(x!)^{1/x}}{x} \cdot x f'(x) \sim \frac{1}{e} \cdot 1 = \frac{1}{e}$$

as $x \to \infty$. Finally, by the mean value theorem,

$$ \lim_{x\to\infty} [(x+1)!^{\frac{1}{x+1}} - x!^{\frac{1}{x}}] = \lim_{x \to \infty} [(x!)^{1/x}]' = \frac{1}{e}. $$

Answer 2

This one is pretty standard. Let us use the fact that $$F(x) =(\Gamma(x+1))^{1/x}/x\to 1/e\tag{1}$$ and note that expression under limit can be written as $$G(x+1)-G(x)=G(x)\left(\frac{G(x+1)}{G(x)}-1\right)$$ where $G(x) =xF(x) $.

Next we observe that $G(x+1)/G(x)\to 1$ and hence using the limit $(t-1)/\log t\to 1$ as $t\to 1$ we can say that desired limit is equal to the limit of $$G(x) \log\frac{G(x+1)}{G(x)}=F(x)\cdot x\log\frac{G(x+1)}{G(x)}$$ The first factor tends to $1/e$ and hence desired limit equals the limit of $$\frac{1}{e}\cdot\log\frac{(G(x+1))^x}{\Gamma(x+1)}=\frac{1}{e}\log\frac{\Gamma(x+2)}{\Gamma(x+1)G(x+1)}=\frac{1}{e}\log\frac{x+1}{G(x+1)}$$ The desired limit is $1/e$ as $G(x+1)/(x+1)=F(x+1)\to 1/e$.

The limit above with $x$ replaced by integer $n$ is already famous on this website and I wasn't sure if same technique can be used for continuous variable $x$ as well. Also it is important to establish the limit in equation $(1)$. This is based on the following lemma.

Lemma: If $f:(a, \infty) \to\mathbb{R} $ is a function such that $f(x+1)-f(x)\to L$ as $x\to\infty$ and further $f$ is bounded on any bounded interval then $f(x) /x\to L$ as $x\to\infty$.

For our current problem we need to take $f(x) =x\log x-\log\Gamma(x+1)$ so that $$f(x+1)-f(x)=x\log\frac{x+1} {x} $$ which tends to $1$ so that $f(x) /x \to 1$ and this is equivalent to equation $(1)$. Further $f$ is bounded on any bounded subset of $(1,\infty)$ so the conditions for above lemma hold.