How to evaluate $\lim\limits_{x\to \infty} x \left(1-\left(\frac{x!}{x^x}\right)^{\frac{1}{x(x+1)} }\right) $?

Question

In this question I tried to solve:

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

I then "reduced" the question to finding: $$\lim\limits_{x\to \infty} x \left(1-\left(\frac{x!}{x^x}\right)^{\frac{1}{x(x+1)} }\right) $$ Here I use $x!$ instead of $\Gamma(x+1)$ because it looks cleaner.

Sangchul Lee Gave a very clever solution to $L:=\lim\limits_{x\to \infty }\Gamma(x+2)^{\frac{1}{x+1}}-\Gamma(x+1)^{\frac{1}{x} }$ However I wonder how one can prove $\lim\limits_{x\to \infty} x \left(1-\left(\frac{x!}{x^x}\right)^{\frac{1}{x(x+1)} }\right)=1 $ without using l'lhopital.

This limit is interesting because it involves $\left(\frac{x!}{x^x}\right)^{\frac{1}{x} }$ and this famous limit is $1/e$.

Answer 1

Here $o(f(x))$ means a function $\alpha(x)$, such that $\lim\limits_{x\to \infty}\frac{\alpha(x)}{f(x)} = 0$. There will be one place in the solution, where I am using the same notation for $\to 0$, but I explicitly state that there.

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Since $x!\sim \sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}$, meaning $x! = \sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}(1+o(1))$, then we have $$\left(\frac{x!}{x^{x}}\right)^{\frac{1}{x(x+1)}} = (\sqrt{2\pi x})^{\frac{1}{x(x+1)}}e^{-\frac{1}{x+1}}(1 + o(1))^{\frac{1}{x(x+1)}} = \exp\left(\frac{\ln(2\pi x)}{2x(x+1)} - \frac{1}{x+1}\right) + \frac{o(1)}{x(x+1)}$$ Note that $e^{-t} = 1-t + o(t),\;t\to 0$, so $$1-\left(\frac{x!}{x^{x}}\right)^{\frac{1}{x(x+1)}} = -\frac{\ln(2\pi x)}{2x(x+1)} + \frac{1}{x+1} + \frac{o(1)}{x(x+1)} + o\left(\frac{\ln(2\pi x)}{2x(x+1)} - \frac{1}{x+1}\right)$$ Also, $$\left|\frac{\ln(2\pi x)}{2x(x+1)} - \frac{1}{x+1}\right| = \frac{|2x-\ln(2\pi x)|}{2x(x+1)}\leq \frac{2x+3x}{2x(x+1)} = \frac{5}{2(x+1)} = o\left(\frac{1}{x}\right),$$ so $$x\left(1-\left(\frac{x!}{x^{x}}\right)^{\frac{1}{x(x+1)}}\right) = -\frac{\ln(2\pi x)}{2(x+1)} + \frac{x}{x+1} + \frac{o(1)}{x+1} + o(1)\stackrel{x\to\infty}{\to} 1$$

Answer 2

I think that it is easier to stay with the problem $$\Gamma(x+2)^{\frac{1}{x+1}}-\Gamma(x+1)^{\frac{1}{x} }$$

Consider $$A(a)=\Gamma(x+a+1)^{\frac{1}{x+a}}$$ Take the logarithm and use Stirling approximation to get after the long division $$\log(A(a))=\log(x)-1+\frac{2 a+\log (2 \pi x)}{2 x}+O\left(\frac{\log (x)}{x^2}\right)$$ Exponentiate again since $A(a)=e^{\log(A(a))}$ to obtain $$A(a)=\frac x e+\frac{2 a+\log (2 \pi x)}{2 e}+O\left(\frac{\log (x)}{x}\right)$$

Apply it twice to get a simple result.

Edit

For the question, consider first $$A=\frac{\Gamma(x+1)}{x^x} \quad \implies \quad \log(A)=\text{log$\Gamma $}(x+1)-x \log (x)$$ Using Striling approximation $$\log(A)=-x+\frac{1}{2} (\log (x)+\log (2 \pi ))+\frac{1}{12 x}+O\left(\frac{1}{x^3}\right)$$

$$\frac 1{x(x+1)} \log(A)=-\frac 1 x+\frac{\log (2 \pi x)+2}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\exp\Bigg(\frac 1{x(x+1)} \log(A) \Bigg)=1-\frac 1 x+\frac{\log (2 \pi x)+3}{2 x^2}+O\left(\frac{1}{x^3}\right)$$

$$x \left(1-\left(\frac{x!}{x^x}\right)^{\frac{1}{x(x+1)} }\right)=1-\frac{\log (2 \pi x)+3}{2 x}+O\left(\frac{1}{x^2}\right)$$

Answer 3

Your expression is $x(1-f(x))$ with $f(x) \to 1$ and hence we can just evaluate the limit of $$-x\log f(x) =x\log(1/f(x))=\frac{1}{x+1}\log\frac{x^x}{x!}=\frac{x}{x+1}\log\frac{x}{(x!)^{1/x}}$$ and this tends to $1\log e=1$.