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It can be shown that $\displaystyle \lim_{n \to \infty}\sqrt[n+1]{(n+1)!}- \sqrt[n]{n!}= \dfrac{1}{e}.$ If we let $f(x)= (\Gamma(1+x))^{\frac{1}{x}}$ and note that $f$ is differentiable over $(0, \infty)$ then by applying the Mean Value Theorem to $f$ over $(n+1, n+2) \, \forall \, n \in \mathbb{N}\cup \{0\}$ we can obtain a sequence $(x_n)$ with $x_n \in (n+1, n+2)$ such that

$\displaystyle \sqrt[n+1]{(n+1)!}- \sqrt[n]{n!} = f'(x_n)$

$\Rightarrow \displaystyle \lim_{n \to \infty} f'(x_n) = \dfrac{1}{e}$

$\Rightarrow \displaystyle \lim_{n \to \infty} f(x_n)\left(\dfrac{x_n \psi(1+x_n)- \ln(\Gamma(1+x_n))}{x_n^{2}}\right) = \dfrac{1}{e}$

$\Rightarrow \displaystyle \lim_{n \to \infty}\dfrac{f(x_n)}{x_n} \displaystyle \lim_{n \to \infty} \left( \dfrac{x_n \psi(1+x_n)- \ln(\Gamma(1+x_n))}{x_n}\right)=\dfrac{1}{e}$

Now $\displaystyle \lim_{n \to \infty}\dfrac{f(x_n)}{x_n} =\dfrac{1}{e}$ since $\displaystyle \lim_{n \to \infty} \dfrac{(n!)^{\frac{1}{n}}}{n} = \dfrac{1}{e}$

$\Rightarrow \displaystyle \lim_{n \to \infty} \left( \dfrac{x_n \psi(1+x_n)- \ln(\Gamma(1+x_n))}{x_n}\right) =1$

$\Rightarrow \displaystyle \lim_{n \to \infty} \left(\psi(1+x_n)-\ln f(x_n)\right)=1$

In the last equality, since the LHS is tending to $1$ through a particular and not generic sequence $(x_n)$ two questions arise:

1) Can the last limit be simplified further to yield some non-trivial limit identity?

2) Can a property of the gamma or digamma function be used here to proceed further?

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The asymptotic expansions of the log Gamma and digamma functions are given respectively by

$$\frac1x \log(\Gamma(x))=\log(x)-1+O\left(\frac1x\right) \tag 1$$

and

$$\psi(x)=\log(x)+O\left(\frac1x\right) \tag2$$

Therefore, we can write

$$\begin{align} \psi(1+x_n)-\frac{\log(\Gamma(x_n))}{x_n}&=\log(1+x_n) -\log(x_n) +1+O\left(\frac1{x_n}\right) \\\\ &=1+O\left(\frac1{x_n}\right) \\\\ &\to 1\,\,\text{as}\,\,n\to \infty \end{align}$$

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