Compute $\lim_{n\to\infty}(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}})$ given $(a_{n+1}-a_n)/n\to a$ and $b_{n+1}/(nb_n)\to b$ 
Let $(a_n)_{n\geq 1}$ and $(b_n)_{n\geq 1}$ be positive real sequences such that
$$\lim_{n\to\infty}\frac{a_{n+1}-a_n}n=a\in \mathbb R_{>0}\qquad\text{and}\qquad \lim_{n\to\infty}\frac{b_{n+1}}{nb_n}=b\in \mathbb R_{>0}.$$
Compute
$$\lim_{n\to\infty}\left(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}}\right).$$

This is W5 of József Wildt International Mathematical Competition, 2020. I have made some progresses and got stuck.
Using Cauchy's criterion, I have obtained that $\sqrt[n]{\frac{b_{n+1}}{b_1n!}}\to b$ and thus
$$\sqrt[n]{\frac{b_n}{b_1n!}}=\sqrt[n]{\frac{b_{n+1}}{b_1n!}}\cdot\sqrt[n]{\frac{nb_n}{b_{n+1}}}\cdot\sqrt[n]{\frac1n}\to b,\qquad\text{as}\ \  n\to\infty.$$
Stirling's formula implies that $\lim_{n\to\infty}\frac{\sqrt[n]{b_n}}n=\frac be$. Now it seems resonable to guess that
$$\lim_{n\to\infty}\left(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}}\right)=\lim_{n\to\infty}\left(\frac{\frac{a_{n+1}}{n}}{\frac{\sqrt[n+1]{b_{n+1}}}{n}}-\frac{\frac{a_n}n}{\frac{\sqrt[n]{b_n}}n}\right)\overset{?}{=}\lim_{n\to\infty}\frac eb\frac{a_{n+1}-a_n}n=\frac{ae}b\tag{1};$$
or another possibility
$$\lim_{n\to\infty}\left(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}}\right)=\lim_{n\to\infty}\left(\frac{\frac{a_{n+1}}{n+1}}{\frac{\sqrt[n+1]{b_{n+1}}}{n+1}}-\frac{\frac{a_n}n}{\frac{\sqrt[n]{b_n}}n}\right)\overset{?}{=}\frac eb\lim_{n\to\infty}\left(\frac{a_{n+1}}{n+1}-\frac{a_n}n\right),\tag{2}$$
where the limit $\lim_{n\to\infty}\left(\frac{a_{n+1}}{n+1}-\frac{a_n}n\right)$ can be obtained by
$$\frac{a_{n+1}}{n+1}-\frac{a_n}n=\frac{a_{n+1}-a_n}n+\frac{a_{n+1}}{n+1}-\frac{a_{n+1}}n=\frac{a_{n+1}-a_n}n-\frac{a_{n+1}}{n(n+1)}$$
and Cesaro-Stolz: $\lim_{n\to\infty}\frac{a_{n+1}}{n(n+1)}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{n(n+1)-(n-1)n}=\frac a2$, so $\lim_{n\to\infty}\left(\frac{a_{n+1}}{n+1}-\frac{a_n}n\right)=a-\frac a2=\frac a2$. In $(1)$ and $(2)$ above, I don't know the eact reasons for both '$?$', I just guess them.
Therefore, when I looked at the same limit from two different points of view, namely $(1)$ and $(2)$, I got different results, which makes me confused. I wonder which one is correct and what is the rigorous proof.
Could someone help me to finish this problem?
 A: Let $c_n=a_n/\sqrt[n] {b_n} $ and then we have $$\frac{c_{n+1}}{c_n}=\frac{a_{n+1}}{a_n}\cdot\frac{\sqrt[n]{b_n}/n}{\sqrt[n+1]{b_{n+1}}/(n+1)}\cdot\frac{n}{n+1}$$ so that the ratio $c_{n+1}/c_n$ tends to same limit as that of $a_{n+1}/a_n$.
Using Cesaro-Stolz we see that $2a_n/n^2\to a$. And hence $a_{n+1}/a_n\to 1$ and therefore $c_{n+1}/c_n\to 1$.
We can next observe that $$c_{n+1}-c_n=c_n\cdot\left(\frac{c_{n+1}}{c_n}-1\right)=\frac{a_n/n} {\sqrt[n] {b_n} /n} \cdot\frac{c_{n+1}/c_n-1}{\log(c_{n+1}/c_n)}\cdot\log\frac{c_{n+1}}{c_n}$$ And then desired limit is same as that of $$\frac{e} {b}\cdot \frac{a_n} {n^2}\cdot n\log\frac{c_{n+1}}{c_n}$$ The limit of above expression is same as that of $$\frac{ae} {2b} \left(n\log\frac{a_{n+1}}{a_n}+\log\frac{b_n\sqrt[n+1]{b_{n+1}}}{b_{n+1}}\right) $$ The second term in parentheses can be written as $$\log\left(\frac{nb_n} {b_{n+1}}\cdot\frac{\sqrt[n+1]{b_{n+1}}}{n+1}\cdot\frac{n+1}{n}\right) $$ and this tends to $$\log((1/b)(b/e)(1))=-1$$ and the first term can be written as $$n\cdot\frac{\log (a_{n+1}/a_n)} {(a_{n+1}/a_n)-1}\cdot\left(\frac{a_{n+1}}{a_n}-1\right)$$ whose limit is same as that of $$\frac{n^2}{a_n}\cdot\frac{a_{n+1}-a_n}{n}$$ which is $(2/a) \cdot a=2$. The desired limit in question is thus $\dfrac{ae} {2b} $.
