How find this sequence $\{x_{n}\}$ such $\lim_{n\to \infty}x_{n}\left(1-\frac{n(1-na_{n})}{\ln{n}}\right)=1$ Consider the sequence $\{a_{n}\}$ satisfying $a_{1}\in(0,1)$,such
$$a_{n+1}=a_{n}(1-a_{n})$$
question:
Find a sequence $\{x_{n}\}$ such that
$$\lim_{n\to \infty}x_{n}\left(1-\dfrac{n(1-na_{n})}{\ln{n}}\right)=1$$
I have prove this
$$\lim_{n\to \infty}\dfrac{n}{\ln{n}}(1-na_{n})=1$$
But I can't find this $x_{n}$  Thank you
such as:How prove this $\displaystyle\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\frac{n}{\ln{n}}\right)=-1$
 A: This is a question of determining the behavior of the solutions to
$$
a_{n+1}=a_{n}(1-a_{n})
$$
as an asymptotic expansion as $n\rightarrow\infty$.  I'll reiterate how to derive the result you already have, then find the next term.  First, note that
$$
\Delta a_n\equiv a_{n+1}-a_{n}=-a_n^2;
$$
this is analogous to the differential equation $a'(n)=-a(n)^2$, whose general solution is
$$
a(n)=\frac{1}{n+C},$$ suggesting that $a_n \sim 1/n$ for large $n$.  Defining an auxiliary sequence $b_n$ via
$$
a_n=\frac{1}{n}\left(1+b_n\right),
$$
and substituting this into the equation for $\Delta a_n$, we find (after some algebra) the form of $\Delta b_n$ to be
$$
\Delta b_{n}=-\frac{1}{n^2}-\frac{1}{n}\left(1+\frac{2}{n}\right)b_n - \frac{1}{n}\left(1+\frac{1}{n}\right)b_n^2.
$$
The leading-order terms are analogous to $b'(n)=-1/n^2-b(n)/n$, whose solution is $b(n)=(A/n -\log n/n)$, which behaves as $-\log n/n$ for large $n$.  Indeed, taking $b_n\sim -\log n/n$ makes both sides of the above equation asymptotic to $\log n/n^2$.  We repeat the procedure of defining a new variable:
$$
b_n=-\frac{\log n}{n} + c_n.
$$
Note that
$$
1-\frac{n(1-na_n)}{\log n}=1+\frac{nb_n}{\log n}=\frac{nc_n}{\log n},
$$
so the dominant behavior of $c_n$ is all we need.  Plugging in the definition, we get
$$
\Delta c_n = \Delta\left(\frac{\log n}{n}\right)-\frac{1}{n^2}+\frac{1}{n}\left(1+\frac{2}{n}\right)\left(\frac{\log n}{n}-c_n\right)-\frac{1}{n}\left(1+\frac{1}{n}\right)\left(\frac{\log n}{n}-c_n\right)^2.
$$
Note that
$$
\Delta\left(\frac{\log n}{n}\right)=\frac{\log (n+1)}{n+1} - \frac{\log n}{n}=\frac{n\log(n+1)-(n+1)\log n}{n(n+1)} \\
= -\frac{\log n}{n^2} + \frac{1}{n^2} + \frac{\log n}{n^3} - \frac{3}{2n^3} + \ldots;
$$
the first two terms cancel the $-1/n^2$ and $\log n/n^2$ terms in the remainder of the right-side, as expected.  We are left with
$$
\Delta c_n = \left(\frac{\log n}{n^3}+\ldots\right) + \left(-\frac{c_n}{n}+\frac{2\log n}{n^3}+\ldots\right) +\left(-\frac{\log^2 n}{n^3}+\ldots\right).
$$
The leading-order equation is just $\Delta c_n \sim -c_n / n$, with solution $c_n \sim K/n$.  Plugging everything back in, we find that
$$
1-\frac{n(1-na_n)}{\log n} \sim \frac{K}{\log n},
$$
and so
$$
\lim_{n\rightarrow\infty}\left(\log n\right)\left(1-\frac{n(1-na_n)}{\log n}\right)=K(a_0);
$$
i.e., the limit exists for each starting value $a_0\in(0,1)$.  However, despite how your question is written, the value of the limit is not universal; it depends on the starting value.  Empirically, the form of the dependence looks something like
$$
K(a_0)\approx -\frac{\log^{4/5} a_1}{2.9a_1},
$$
where $a_1=a_0(1-a_0)$.
