# Is the function $f(x)=\sqrt[x+1]{\Gamma(x+2)}-\sqrt[x]{\Gamma(x+1)}$ decreasing?

We know that the limit of $$a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}$$ is $$1/e=0.3678794411714423215955237701614608674458111310317678345078368016$$, see What is $\lim_{n\to \infty }\left(\sqrt[\leftroot{-2}\uproot{2}n+1]{(n+1)!}-\sqrt[\leftroot{-2}\uproot{2}n]{n!}\right)$?

Use Wolfram we have that $$a_1=\sqrt[2]{2!}-\sqrt[1]{1!}=0.4142135623730950488016887242096980785696718753769480731766797379$$ $$a_2=\sqrt[3]{3!}-\sqrt[2]{2!}=0.4029070304590446100895230321175624238585385877642715983046545599$$ $$a_3=\sqrt[4]{4!}-\sqrt[3]{3!}=0.3962432465685035259263687905534027946796259019185992795523061114$$ $$a_4=\sqrt[5]{5!}-\sqrt[4]{4!}=0.3918072452967087075081863770479530310194707973034612596596597742$$,... an so on. From this data we can guess that $$a_n$$ is decreasing. More generally, let $$f(x)=\sqrt[x+1]{\Gamma(x+2)}-\sqrt[x]{\Gamma(x+1)},$$ where $$\Gamma(x)$$ is the Gamma function $$\Gamma(x)=\int_0^\infty e^{-t}t^x\dfrac{dt}{t}.$$ Plot the graph of $$f(x)$$, we can see that $$f$$ decreases. Can anyone prove this conclusion？

• I think is is sufficient to prove that $g(x) = \Gamma(x+1)^{\frac1x}$'s derivative is decreasing. Commented Jul 9, 2021 at 4:39
• @Trebor Yes, $\frac{d}{dx}(\sqrt[x]{\Gamma(x+1)})=\sqrt[x]{\Gamma(x+1)} \cdot\frac{x \psi(1 + x)-\log(\Gamma(1 + x))}{x^2}$ where $\psi(x)=\frac{d}{dx}\ln\big(\Gamma(x)\big)=\frac{\Gamma'(x)}{\Gamma(x)}$ is the digamma function. The derivative of $\Gamma(x+1)^{\frac{1}{x}}$ when $x\to\infty$ is zero and if we can prove the derivative is decreasing, then we will prove it. But how to prove the derivative is decreasing, the second derivative of $g(x)$ is very complicated.
– HGF
Commented Jul 9, 2021 at 5:33
• I was thinking of a way to cleverly use the log-concavity of $\Gamma(x)$, but apparently it doesn't work. Commented Jul 9, 2021 at 5:35
• There is an article : Inequalities for the Gamma and Polygamma Functions By H. ALZER link.springer.com/article/10.1007/BF02942573 Maybe it is useful.
– HGF
Commented Jul 9, 2021 at 5:43

Sketch of a proof:

Let $$f(x) = \Gamma(x + 1)^{1/x}$$.

We have $$f''(x) = \frac{\Gamma(x + 1)^{1/x}}{x^2} \left[x\psi'(x + 1) - 1 + \left(\frac{\ln \Gamma(x + 1)}{x} + 1 - \psi(x + 1)\right)^2\right]$$ where $$\psi(\cdot)$$ is the digamma function defined by $$\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$$.

Fact 1: $$f''(x) < 0$$ for all $$x \ge \frac25$$. (The proof is given at the end.)

Fact 2: $$f''(x) < 0$$ for all $$0 < x < \frac25$$. (The proof is given at the end.)

From Facts 1-2, we have $$f''(x) < 0$$ for all $$x > 0$$, and the desired result follows.

We are done.

Proof of Fact 1:

It suffices to prove that, for all $$x \ge \frac25$$, $$x\psi'(x + 1) - 1 + \left(\frac{\ln \Gamma(x + 1)}{x} + 1 - \psi(x + 1)\right)^2 < 0.$$

Fact 3: $$\psi(u) > \ln u - \frac{1}{2u} - \frac{1}{12u^2}$$ for all $$u > 0$$.
(See: Theorem 5, [1].)

Fact 4: $$\psi'(u) < \frac{1}{u} + \frac{1}{2u^2} + \frac{1}{6u^3}$$ for all $$u > 0$$.
(See: Theorem 4, [1].)

Fact 5: $$\Gamma(x) \le \sqrt{2\pi}\, x^{x - 1/2}\mathrm{e}^{-x} \mathrm{e}^{\frac{1}{12x}}$$ for all $$x > 0$$.

Fact 6: $$\frac{\ln \Gamma(x + 1)}{x} + 1 > \psi(x + 1)$$ for all $$x > 0$$.
(The proof is given at the end.)

By using Facts 3-6, it suffices to prove that \begin{align*} &x\cdot \left(\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{6(x + 1)^3}\right) - 1 \\ & + \bigg[\frac{\ln \left(\sqrt{2\pi} (x + 1)^{x + 1/2}\mathrm{e}^{-x - 1} \mathrm{e}^{\frac{1}{12(x + 1)}}\right)}{x} + 1 - \bigg(\ln (x + 1) - \frac{1}{2(x + 1)} - \frac{1}{12(x + 1)^2}\bigg)\bigg]^2\\ &< 0 \end{align*} or \begin{align*} &\ln \left[\sqrt{2\pi} (x + 1)^{x + 1/2}\mathrm{e}^{-x - 1} \mathrm{e}^{\frac{1}{12(x + 1)}}\right] + x - x\bigg(\ln (x + 1) - \frac{1}{2(x + 1)} - \frac{1}{12(x + 1)^2}\bigg)\\ &< x\sqrt{\frac{3x^2 + 8x + 6}{6(x + 1)^3}} \end{align*} or \begin{align*} \ln \sqrt{2\pi} - 1 + \frac{1}{2}\ln(x + 1) + \frac{6x^2 + 8x + 1}{12(x + 1)^2} - x\sqrt{\frac{3x^2 + 8x + 6}{6(x + 1)^3}} < 0. \end{align*} Denote LHS by $$g(x)$$. We have $$g'(x) = \frac{3x^2 + 8x + 6}{6(x + 1)^3} - {\frac {3\,{x}^{3}+12\,{x}^{2}+18\,x+12}{6\,{x}^{3}+22\,{x}^{2}+28\,x+ 12}} \sqrt{\frac{3x^2 + 8x + 6}{6(x + 1)^3}}.$$ It is easy to prove that $$g'(x) < 0$$ for all $$x > 0$$. Also, $$g(2/5) < 0$$. Thus, $$g(x) < 0$$ for all $$x \ge 2/5$$.

We are done.

Proof of Fact 6:

It suffices to prove that, for all $$x > 0$$, $$\ln \Gamma(x + 1) + x - x \psi(x + 1) > 0.$$ Denote LHS by $$h(x)$$. We have $$h'(x) = 1 - x\psi'(x+1) > 1 - x \cdot \left(\frac{1}{x + 1} + \frac{1}{(x + 1)^2}\right) > 0$$ where we have used $$\psi'(u) \le \frac{1}{u} + \frac{1}{u^2}$$ for all $$u > 0$$. Also, $$h(0) = 0$$. Thus, $$h(x) > 0$$ for all $$x > 0$$.

We are done.

Proof of Fact 2:

It suffices to prove that, for all $$0 < x < \frac25$$, $$x\psi'(x + 1) - 1 + \left(\frac{\ln \Gamma(x + 1)}{x} + 1 - \psi(x + 1)\right)^2 < 0.$$

Fact 9: $$\psi'(x + 1) \le \sum_{k=0}^6 \frac{x^k}{k!}\psi^{(k + 1)}(1)$$ for all $$x > 0$$.
(Remarks: I will add the proofs of Facts 9-11 in the future.)

Fact 10: $$\psi(x + 1) \ge \sum_{k=0}^6 \frac{x^k}{k!}\psi^{(k)}(1)$$ for all $$x > 0$$.

Fact 11: $$\ln\Gamma(x + 1) \le \sum_{k=1}^6 \frac{x^k}{k!}\psi^{(k - 1)}(1)$$ for all $$x > 0$$.

By using Facts 9-11 and Fact 6, it suffices to prove that \begin{align*} x \sum_{k=0}^6 \frac{x^k}{k!}\psi^{(k + 1)}(1) - 1 + \left(\sum_{k=0}^5 \frac{x^k}{(k + 1)!}\psi^{(k)}(1) + 1 - \sum_{k=0}^6 \frac{x^k}{k!}\psi^{(k)}(1)\right)^2 < 0 \end{align*} that is $$\frac{x^2}{514382400}(q_{10}x^{10} + q_9 x^9 + \cdots + q_1x + q_0) < 0 \tag{1}$$ where $$q_{10} = 514382400[\zeta(7)]^2, \cdots, q_0 = 3572100\pi^4 - 342921600\zeta(3)$$. One can prove that (1) holds for all $$0 < x < 2/5$$.

We are done.

References

[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.

You could even make the problem more general $$f_n(x)=\sqrt[x+n]{\Gamma(x+n+1)}-\sqrt[x+n-1]{\Gamma(x+n)}$$ $$A=\sqrt[x+n]{\Gamma(x+n+1)}\implies \log(A)=\frac 1{x+n}\log(\Gamma(x+n+1))$$ Now, use Stirling approximation and continue with Taylor series to get $$\log(A)$$ and Taylor series again using $$A=e^{\log(A)}$$.
Do the same for the second piece and you will end with $$f_n(x)=\frac 1e \left(1+\frac 1{2x}+\cdots\right)$$ The contribution of $$n$$ will appear in the $$\cdots$$