Evaluate $\lim\limits_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$

I think I completed a proof thanks to Gary's idea with the $c_n$ substitution and the ratio limits. Bear with me. First, let's make a little different substitution $c_n = \sqrt[n]{\dfrac{a_n}{b_n}}$ and use the following criterion.

Lemma (Cauchy-d'Alembert): If $x_n > 0,\ (\forall) n \in \mathbb{N}$ and the limit $l = \lim\limits_{n\to \infty} \dfrac{x_{n+1}}{x_n}$ exists, then $\sqrt[n]{x_n}$ converges to $l$.

Since:

$$\lim_{n \to \infty}\frac{\dfrac{a_{n+1}}{(n+1)^{n+1}\cdot b_{n+1}}}{\dfrac{a_n}{n^n\cdot b_n}} = \lim_{n \to \infty} \left(\frac{a_{n+1}}{n^2\cdot a_n} \cdot \frac{n\cdot b_n}{b_{n+1}} \cdot \frac{n^{n+1}}{(n+1)^{n+1}}\right) = \frac{x}{y \cdot e}$$

we can apply Cauchy D'Alembert for

$$\dfrac{c_n}{n} = \sqrt[n]{\frac{a_n}{n^n\cdot b_n}} \to \frac{x}{y \cdot e}$$

From this we also have

$$\lim\limits_{n\to\infty}\, \frac {c_{n+1}}{c_n}=\lim\limits_{n\to\infty}\, \left[\frac {c_{n+1}}{n+1}\cdot\frac n{c_n}\cdot\frac {n+1}n\right]=1$$

and

$$\lim\limits_{n\to\infty}\, \left(\frac {c_{n+1}}{c_n}\right)^n=\lim\limits_{n\to\infty}\, \left(\frac {a_{n+1}}{n^2\cdot a_n}\cdot\frac {n\cdot b_n}{b_{n+1}}\cdot n \cdot \sqrt[n]{\frac {b_n}{a_n}}\right)^{\frac n{n+1}}=\frac xy\cdot\frac {y\cdot e}x=e$$

Therefore, we can compute the required limit as follows:

$$ \begin{aligned} \lim_{n\to\infty}\, \left(\sqrt[n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right) &= \lim_{n\to\infty}\, \left(c_{n+1}-c_n\right) \\ &=\lim_{n\to\infty}\, \left[c_n\cdot\left(\frac {c_{n+1}}{c_n}-1\right)\right]\\ &= \lim_{n\to\infty}\ \left[\frac {c_n}n\cdot\frac {e^{\ln\left(\dfrac {c_{n+1}}{c_n}\right)}-1}{\ln\left(\dfrac {c_{n+1}}{c_n}\right)}\cdot\ln\left(\frac {c_{n+1}}{c_n}\right)^n\right] \\ &= \frac x{y\cdot e}\cdot 1\cdot\ln e \\ &=\frac x{y\cdot e} \end{aligned} $$

This completes the solution. And interesting remark is that choosing $a_n= (n!)^2$ and $b_n = n!$, we get the following limit (Lalescu):

$$\lim_{n\to\infty}\, \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\frac 1{e}$$

Some observations. Define $$ c_n : = \frac{1}{n!}\frac{a_n}{b_n}. $$ Then by the assumptions on $a_n$ and $b_n$, we have $$ \mathop {\lim }\limits_{n \to + \infty } \frac{c_{n + 1}}{c_n} = \frac{x}{y}\quad \text{ and hence }\quad \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{c_n} = \frac{x}{y}. $$ Note that $$ \sqrt[n]{\frac{a_n}{b_n}} = \sqrt[n]{n!c_n}. $$ By Stirling's formula $$ \sqrt[n]{{n!}} = \frac{n}{e} + \frac{{\log (2\pi n)}}{{2e}} + \mathcal{O}\left( {\frac{1}{n}} \right). $$ If it was true that $$ \sqrt[n]{{c_n }} = \frac{x}{y} +\frac{K}{n}+ o\left( {\frac{1}{n}} \right) $$ then we would have $$ \sqrt[{n + 1}]{{(n + 1)!c_{n + 1} }} - \sqrt[n]{{n!c_n }} = \frac{x}{y}\frac{1}{e} + \mathcal{O}\left( {\frac{1}{n}} \right), $$ meaning that the limit is indeed $\frac{x}{y}\frac{1}{e}$. In any case, $$ \mathop {\lim \inf }\limits_{n \to + \infty } \left( {\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}} \right) = \mathop {\lim \inf }\limits_{n \to + \infty } \frac{{\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}}}{{n + 1 - n}} \le \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt[n]{{n!c_n }}}}{n} = \frac{x}{{ye}} \\ \le \mathop {\lim \sup }\limits_{n \to + \infty } \frac{{\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}}}{{n + 1 - n}} = \mathop {\lim \sup }\limits_{n \to + \infty } \left( {\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}} \right). $$ Thus, if the limit in question exists, it must be $\frac{x}{y}\frac{1}{e}$.