Let $(a_n)_{n \geq 1 } $ be an unbounded, positive sequence of real numbers, such that $\lim\limits_{n \to \infty} \dfrac{a_{n+1}}{na_n}=a, $ then
$$\lim\limits_{n \to \infty}n\left(\frac{{a_{n+1}}^{\frac{1}{n}}}{n+1}-\frac{{a_{n}}^{\frac{1}{n}}}{n}\right)=0$$
Of course, the sequence being unbounded and of positive, real, numbers, $a > 0$, and I have managed to obtain, from Cauchy's Criterion, that $ \lim\limits_{ n \to \infty} (\frac{{a_n}}{n!})^{\frac{1}{n}}=a, $ and, further, by using Stirling's limit, obtaining $ \lim\limits_{ n \to \infty} \frac{{a_n}^{\frac{1}{n}}}{n}=a $ is quite straight-forward. A particular example, which verifies the given property, $a_n= a^n n!,$ indeed verifies the limit-to-be-proven.
This is all (which is relevant) that I've managed to obtain regarding this problem, any help is welcome.
$$ \frac {a_{n+1}}{na_n} = \frac {a_{n+1} (n-1)!} {a_{n} n!} = \frac {a_{n+1}/n!} {a_{n} / (n-1)!} \sim a \quad [n \to +\infty]. $$
Let $b_n = a_n/(n-1)!$, then $b_{n+1} / b_n \sim a$. For $\varepsilon > 0$, there is some $N \in \Bbb N^*$ that $a + \varepsilon > b_{n+1}/b_n > a - \varepsilon $ whenever $n \geqslant N$. WLOG we just assume that $N = 1$. Then $b_{n+1}/b_1 \in ((a- \varepsilon)^n, (a+\varepsilon )^n)$, which means $(a_{n+1}/(a_1n!))^{1/n} \in (a - \varepsilon, a+ \varepsilon) $. Thus
$$ a_{n+1} \sim a^n a_1 n!. $$ Now the target
\begin{align*} & \qquad n \left( \frac {a_{n+1}^{1/n}} {n+1} - \frac {a_n^{1/n}} n\right)\\ &= n \cdot \frac {a_n^{1/n}} n \left( \left( \frac {a_{n+1}} {a_n}\right)^{1/n} \cdot \frac n{n+1} -1\right) \\ &= a_n^{1/n} \left( \left( \frac {a_{n+1}} {a_n}\right)^{1/n} \cdot \frac n{n+1} -1\right) \end{align*} Since
\begin{align*} &\qquad \left(\frac {a_{n+1}} {a_n}\right)^{1/n} \cdot \frac n{n+1} -1\\ &= \exp \left( \frac 1n \log \left( \frac {a_{n+1}} {a_n}\right) - \log \left(1 + \frac 1n\right) \right) - 1 \\ &\sim \exp \left( \frac {\log n} n + \frac {\log a}n - \frac 1n + o \left( \frac 1n\right) \right) - 1\\ &\sim \exp \left( \frac {\log n} n + O\left( \frac 1n\right) \right) - 1\\ &\sim \frac {\log n}n, \end{align*} we conclude that \begin{align*} & \qquad n \left( \frac {a_{n+1}^{1/n}} {n+1} - \frac {a_n^{1/n}} n\right)\\ &= a_n^{1/n} \left( \left( \frac {a_{n+1}} {a_n}\right)^{1/n} \cdot \frac n{n+1} -1\right)\\ &\sim a_n^{1/n} \cdot \frac {\log n}n\\ &\sim a^{(n-1)/n} a_1^{1/n} ((n-1)!)^{1/n} \cdot \frac {\log n}n\\ &\sim a (2 \pi (n-1))^{1/(2n-2)} \left(\frac {n-1} {\mathrm e}\right)^{1 - 1/n} \cdot \frac {\log n}n \tag {Stirling}\\ &\sim \frac a{\mathrm e} \log n\\ &\xrightarrow {n \to +\infty} +\infty. \end{align*}
I will leave it here... any discussion is welcomed. But if I was right, then the limit shall be $+\infty$.