If $\frac{p_{n+1}}{np_n} \to p > 0 $, then $\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_{n}} \to \frac{p}{e}$ Problem:
Prove that,
if a sequence
${p_n}$
satisfies
$p_n > 0$
and
$\lim\limits_{n \to \infty}
\frac{p_{n+1}}{np_n} = p > 0
$,
then
$\lim\limits_{n \to \infty}
\left(\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_{n}} \right)
=\frac{p}{e}
$.
This lemma occurs 
in problem B-1151
in the current Fibonacci Quarterly
(August 2015)
with a reference to a paper
published by the
University of Belgrade
and not written in English.
It is used to prove things like
$\lim\limits_{n \to \infty}
\left(\sqrt[n+1]{(n+1)!F_{n+1}}-\sqrt[n]{n!F_{n}} \right)
=\frac{\alpha}{e}
$
where
$\alpha
=\frac{1+\sqrt{5}}{2}
$.
This looks like an interesting result,
so I thought that
I would try to prove it.
Here is my attempt.
We have
$\frac{p_{n+1}}{np_n} 
\to p
$.
If this was
$\frac{p_{n+1}}{(n+1)p_n} 
\to p
$,
this could be divided by
$\frac{n!}{n!}$
to
get
$\frac{p_{n+1}/(n+1)!}{p_n/n!} 
\to p
$.
From this,
setting
$p_n/n! = q_n$,
this would become
$\frac{q_{n+1}}{q_n}
\to p
$,
and things look hopeful.
But,
since
$\frac{n}{n+1}
\to 1$,
$p
=\lim\limits_{n \to \infty}\frac{p_{n+1}}{np_n}
=\lim\limits_{n \to \infty}\frac{n}{n+1} \frac{p_{n+1}}{np_n}
=\lim\limits_{n \to \infty}\frac{p_{n+1}}{(n+1)p_n} 
$.
Proceeding as described,
letting
$p_n/n!
=q_n
$,
$p
=\lim\limits_{n \to \infty} \frac{q_{n+1}}{q_n}
$,
so
$\lim\limits_{n \to \infty} \frac{q_n}{p^n}
= a
$
for some $a > 0$.
Therefore
$p_n
\approx n!p^n a
$.
Since
$(n!)^{1/n}
\to \frac{n}{e}
$
and
$a^{1/n}
\to 1
$,
$\sqrt[n]{p_{n}} 
\approx \sqrt[n]{n!p^n a}
\to \frac{n}{e}p 
$
so
$\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_{n}} 
\to  \frac{n+1}{e}p- \frac{n}{e}p
= \frac{p}{e}
$
and we are done.
My questions are:
(1) Is my prove valid?
(2) Is there are better proof?
(3) Is there a more refined result, with more terms beyond
$\frac{p}{e}$?
 A: Your argument doesn't quite work, unfortunately. From $\lim\limits_{n\to \infty} \frac{q_{n+1}}{q_n} = p$, it doesn't follow that $\lim\limits_{n\to\infty} \frac{q_n}{p^n}$ exists, or if it exists, that it is a positive real number. Consider for example $q_n = n^{\alpha}$. Then we have $\frac{q_{n+1}}{q_n} \to 1$ for all $\alpha\in \mathbb{R}$, but $q_n \to 0$ for $\alpha < 0$ and $q_n \to +\infty$ for $\alpha > 0$. For an example where the limit doesn't exist in the extended sense, consider a sequence $(q_n)$ that oscillates between say $2$ and $3$, with $\lvert q_{n+1} - q_n\rvert \to 0$. Also, from the asymptotic equality $\sqrt[n]{n!} \sim \frac{n}{e}$ alone, we cannot deduce $\sqrt[n+1]{(n+1)!} - \sqrt[n]{n!} \to \frac{1}{e}$. We need more precise information for that result.
However, from $\frac{q_{n+1}}{q_n} \to p$, it follows that $\sqrt[n]{q_n} \to p$, since for positive sequences $(a_n)$ we have the inequalities
$$\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \leqslant \liminf_{n\to\infty} \sqrt[n]{a_n} \leqslant \limsup_{n\to\infty} \sqrt[n]{a_n} \leqslant \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}.$$
Let's define $(\beta_n)$ via
$$\frac{q_{n+1}}{q_n} = p\cdot e^{\beta_{n+1}}.$$
We then have
$$q_n = q_0 \cdot p^n \cdot \prod_{\nu = 1}^n e^{\beta_{\nu}}$$
and
$$\sqrt[n]{q_n} = \sqrt[n]{q_0}\cdot p \cdot \exp \Biggl(\frac{1}{n}\sum_{\nu = 1}^n \beta_{\nu} \Biggr).$$
Define $c_n = \frac{1}{n}\sum_{\nu = 1}^n \beta_{\nu}$. Our assumption $\frac{q_{n+1}}{q_n} \to p$ means that we have $\beta_n \to 0$, and hence also the Cesàro means $c_n$ converge to $0$.
By Stirling's approximation,
$$\log n! = \bigl(n+\tfrac{1}{2}\bigr)\log n - n + \log \sqrt{2\pi} + O(n^{-1}),$$
we get
$$\frac{\log n!}{n} = \log n - 1 + \frac{\log (2\pi n)}{2n} + O(n^{-2})$$
and
\begin{align}
\sqrt[n]{n!} &= \frac{n}{e}\exp\biggl( \frac{\log (2\pi n)}{2n} + O(n^{-2})\biggr)\\
&= \frac{n}{e}\biggl(1 + \frac{\log (2\pi n)}{2n} + O\biggl(\frac{(\log n)^2}{n^2}\biggr)\biggr)\\
&= \frac{n}{e} + \frac{\log (2\pi n)}{2e} + O\biggl(\frac{(\log n)^2}{n}\biggr).
\end{align}
Since $\sqrt[n]{q_0} = \exp \frac{\log q_0}{n} = 1 + \frac{\log q_0}{n} + O(n^{-2})$, we get
$$\sqrt[n]{n!\cdot q_0} = \frac{n}{e} + \frac{\log q_0}{e} + \frac{\log (2\pi n)}{2e} + O\biggl(\frac{(\log n)^2}{n}\biggr),\tag{1}$$
and then, with $p_n = n!\cdot q_n$, we obtain
\begin{align}
\sqrt[n+1]{p_{n+1}} - \sqrt[n]{p_n} &= \sqrt[n+1]{(n+1)!\cdot q_0}\cdot p e^{c_{n+1}} - \sqrt[n]{n!\cdot q_0} \cdot p e^{c_n}\\
&= \sqrt[n+1]{(n+1)!\cdot q_0}\cdot p \bigl(e^{c_{n+1}} - e^{c_n}\bigr) + p e^{c_n}\bigl(\sqrt[n+1]{(n+1)!\cdot q_0} - \sqrt[n]{n!\cdot q_0}\bigr).\tag{$\ast$}
\end{align}
Since $c_{n+1} - c_n = \frac{\beta_{n+1} - c_n}{n+1} \in o(1/n)$, we have
$$e^{c_{n+1}} - e^{c_n} = e^{c_n}\bigl(e^{c_{n+1}-c_n}-1\bigr) = e^{c_n}\biggl(\exp\biggl(\frac{\beta_{n+1} - c_n}{n+1}\biggr) - 1\biggr) \in o(1/n),$$
and with $\sqrt[n+1]{(n+1)!} \sim \frac{n+1}{e}$ it follows that the first summand in $(\ast)$ tends to $0$.
From $(1)$, we obtain
$$\sqrt[n+1]{(n+1)!\cdot q_0} - \sqrt[n]{n!\cdot q_0} = \frac{1}{e} + \frac{\log \bigl(1 + \frac{1}{n}\bigr)}{2e} + O\biggl(\frac{(\log n)^2}{n}\biggr) = \frac{1}{e} + O\biggl(\frac{(\log n)^2}{n}\biggr),$$
and since $c_n \to 0$, the second summand in $(\ast)$ tends to $\frac{p}{e}$, giving
$$\lim_{n\to \infty} \bigl(\sqrt[n+1]{p_{n+1}} - \sqrt[n]{p_n}\bigr) = \frac{p}{e}$$
for positive $p_n$ with $\frac{p_{n+1}}{np_n} \to p \in (0,+\infty)$.

Is there a more refined result, with more terms beyond $\frac{p}{e}$?

That depends on what "more terms beyond $\frac{p}{e}$" means. Of course we could keep some more terms involving $c_n$ and $c_{n+1} - c_n$ explicit. In that sense, we could have a more refined result. But without more information about $(\beta_n)$, those terms wouldn't be useful, I think, since the behaviour of these terms could be more or less arbitrary.
A: In Daniel Fischer's answer, the use of Stirling approximation may be replaced by that of lalescu's limit, which is the following particular case of the theorem to prove:
$$\lim_{n\to\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\frac1e.$$
For this, start at the following point of Daniel's answer:
$$
\sqrt[n+1]{p_{n+1}} - \sqrt[n]{p_n} = \sqrt[n+1]{(n+1)!\,q_0}\,p \bigl(e^{c_{n+1}} - e^{c_n}\bigr) + p e^{c_n}\bigl(\sqrt[n+1]{(n+1)!\,q_0} - \sqrt[n]{n!\,q_0}\bigr)\tag{$\ast$}
$$
and $e^{c_{n+1}}-e^{c_n}\in o(1/n).$
The first summand in $(*)$ tends to $0$ because $(n+1)!\le(n+1)^{n+1}.$
The parenthesis in the second summand may be decomposed:
$$\begin{align}\sqrt[n+1]{(n+1)!\,q_0} - \sqrt[n]{n!\,q_0}&=\sqrt[n+1]{(n+1)!}\left(\sqrt[n+1]{q_0}-\sqrt[n]{q_0}\right)+\sqrt[n]{q_0}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)\\&=O(n)\sqrt[n]{q_0}O(1/n^2)+\sqrt[n]{q_0}\left(\frac1e+o(1)\right)\\
&\to\frac1e
\end{align}$$
and the end of the proof if the same as in Daniel's answer.
A: Yet a much simpler proof, derived from @StefanSolomon 's answer to his related own post.
Let $q_n:=\frac{p_n}{n^n}.$ Since $\frac{q_{n+1}}{q_n}=\frac{p_{n+1}}{np_n}\left(1-\frac1{n+1}\right)^{n+1}\to\frac pe,$
$$\frac{\sqrt[n]{p_n}}n=\sqrt[n]{q_n}\to\frac pe$$
hence
$$w_n:=\frac{p_{n+1}}{p_n\sqrt[n+1]{p_{n+1}}}\sim\frac{p_{n+1}}{np_n}\frac{n+1}{\sqrt[n+1]{p_{n+1}}}\to p\frac ep=e.$$
Now,
$$\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_n}=\sqrt[n]{p_n}\left(\sqrt[n]{w_n}-1\right)\sim\frac{\sqrt[n]{p_n}}n\ln w_n\to\frac pe.$$
