TITLE: Limit related to the recursion $a_{n+1}=(1-n^{-1/4}a_n)a_n$
QUESTION [6 upvotes]: Let $0<a_{1}<1$, with
$$a_{n+1}=a_{n}(1-n^{-\frac{1}{4}}a_{n})$$
Does there a real numbers $A$  which make the limit
$$ \lim_{n\to\infty}\sqrt[4]{n}\left(A-\frac{4}{3}\sqrt[4]{n^3}\left(1-\frac{4}{3}\sqrt[4]{n^3}a_{n}\right) \right)=B$$
exist? if possible find the value of $A$ and $B$.
Here is my approach.
First it is easy to find that $\{a_{n}\}$ is a strict decrease sequcence with $a_{n}\to 0 (n\to+\infty)$,and by O.Stolz Therom,we have:
\begin{align*}
\lim_{n\to\infty}\sqrt[4]{n^3}a_{n}&=\lim_{n\to\infty}\frac{n^{\frac{3}{4}}}{\frac{1}{a_{n}}}\\
&=\lim_{n\to\infty}\frac{(n+1)^{\frac{3}{4}}-n^{\frac{3}{4}}}{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}}\\
&=\lim_{n\to\infty}\frac{n^{\frac{3}{4}}\left(\left(1+\frac{1}{n}\right)^{\frac{3}{4}}-1\right)}{\frac{1}{a_{n}(1-n^{-\frac{1}{4}}a_{n})}-\frac{1}{a_{n}}}\\
&=\lim_{n\to\infty}n\left(\left(1+\frac{1}{n}\right)^{\frac{3}{4}}-1\right)(1-n^{-\frac{1}{4}a_{n}})\\
&=\frac{3}{4}
\end{align*}
which implys that
$$ a_{n}\sim \frac{3}{4}n^{-\frac{3}{4}} $$
Also we can rewrite the condition into
$$ \frac{1}{a_{n+1}}=\frac{1}{a_{n}}+\frac{n^{-\frac{1}{4}}}{1-n^{-\frac{1}{4}}a_{n}}$$
and expand it
$$\frac{1}{a_{n+1}}-\frac{1}{a_{n}}=n^{-\frac{1}{4}}\left(1+n^{-\frac{1}{4}}a_{n}+n^{-\frac{1}{2}}a_{n}^2+o\left(n^{-\frac{1}{2}}a_{n}^2\right)\right)$$
but 
$$ a_{n}\sim \frac{3}{4}n^{-\frac{3}{4}}\Rightarrow \frac{1}{a_{n}}\sim\frac{4}{3}n^{\frac{3}{4}}$$
hence we use O.Stolz again to estimate $\frac{1}{a_{n}}$,it is
\begin{align*}
&\lim_{n\to\infty}\frac{\frac{1}{a_{n}}-\frac{4}{3}n^{\frac{3}{4}}}{n^{-\frac{1}{4}}}\\
&=\lim_{n\to\infty}\frac{\frac{1}{a_{n+1}}-\frac{1}{a_{n}}-\frac{4}{3}\left((n+1)^{\frac{3}{4}}-n^{\frac{3}{4}}\right)}{(n+1)^{-\frac{1}{4}}-n^{-\frac{1}{4}}}\\
&=\lim_{n\to\infty}\frac{n^{-\frac{1}{4}}+n^{-\frac{1}{2}}a_{n}+n^{-\frac{3}{4}}a_{n}^{2}+o(n^{-\frac{3}{4}}a_{n}^{2})-\frac{4}{3}n^{\frac{3}{4}}\left[\left(1+\frac{1}{n}\right)^{\frac{3}{4}}-1\right]}{n^{-\frac{1}{4}}\left[\left(1+\frac{1}{n} \right)^{-\frac{1}{4}}-1\right]}\\
&=\lim_{n\to\infty}\frac{n^{-\frac{1}{2}}a_{n}+\frac{1}{8}n^{-\frac{5}{4}}+n^{-\frac{3}{4}}a_{n}^{2}+o(n^{-\frac{5}{4}})}{-\frac{1}{4}n^{-\frac{5}{4}}}=-\frac{7}{2}
\end{align*}
Which leads 
$$\frac{1}{a_{n}}=\frac{4}{3}n^{\frac{3}{4}}-\frac{7}{2}n^{-\frac{1}{4}}+o(n^{-\frac{1}{4}})$$
\begin{align*}
&a_{n}=\frac{1}{\frac{4}{3}n^{\frac{3}{4}}-\frac{7}{2}n^{-\frac{1}{4}}+o(n^{-\frac{1}{4}})}\\
&=\frac{3}{4}n^{-\frac{3}{4}}\cdot \frac{1}{1-\frac{21}{8}\frac{1}{n}+o(\frac{1}{n})}\\
&=\frac{3}{4}n^{-\frac{3}{4}}\left(1+\frac{21}{8}\frac{1}{n}+o(\frac{1}{n})\right)\\
&=\frac{3}{4}n^{-\frac{3}{4}}+\frac{63}{32}n^{-\frac{7}{4}}+o(n^{-\frac{7}{4}})
\end{align*}
After some simple compute,we find that $A=0$,and $B=\frac{7}{2}$,
The man who propose this problem told me my solution was obviously wrong,but I didn't know where is wrong? Can someone help me to verify it? Or if you have better idea about such problem,I also do appreciate. Thank you very much!

REPLY [3 votes]: Because
$$
a_{n+1}-a_n=-n^{-1/4}a_n^2\tag{1}
$$
$a_n$ is decreasing. Furthermore, if $0\le a_n\le1$, then since
$$
a_n(1-a_n)\le a_n\left(1-n^{-1/4}a_n\right)\le a_n\tag{2}
$$
we have $0\le a_{n+1}\le1$.
Fiddling with $(1)$ and the defining equation, we get
$$
\begin{align}
\frac1{a_{n+1}}-\frac1{a_n}
&=\frac{a_n-a_{n+1}}{a_na_{n+1}}\\[3pt]
&=n^{-1/4}\frac{a_n}{a_{n+1}}\\
&=\frac1{n^{1/4}-a_n}\tag{3}
\end{align}
$$
Using the Euler-Maclaurin Sum Formula, then plugging back into $(3)$ and applying the Euler-Maclaurin Sum Formula again, we get
$$
\begin{align}
\frac1{a_n}
&=\frac43n^{3/4}+O(n^{1/2})\\
&=\frac43n^{3/4}+C-\frac72n^{-1/4}+O(n^{-1/2})\tag{4}
\end{align}
$$
Therefore,
$$
a_n=\frac34n^{-3/4}-\frac{9C}{16}n^{-3/2}+\frac{63}{32}n^{-7/4}+O(n^{-2})\tag{5}
$$
Plugging $(5)$ into
$$
\lim_{n\to\infty}n^{1/4}\left(A-\frac43n^{3/4}\left(1-\frac43n^{3/4}a_n\right) \right)=B\tag{6}
$$
We get $A=C$ and $B=\frac72$, but I am still working on relating $C$ to $a_1$.