Let $b_n=1/a_n$.Obviously $\lim_{n\to\infty}b_n=0,\lim_{n\to \infty }\frac{b_{n+1}}{b_n}=1$,then
\begin{aligned}
\lim_{n\to \infty}\frac{a_n^{1+1/k}}{n}&=\lim_{n\to \infty}\frac{(1/b_n)^{1+1/k}}{n}
\\& =\lim_{n\to \infty}\left(\frac{1}{b_{n+1}^{1+1/k}}-\frac{1}{b_{n}^{1+1/k}}\right) &\text{(Cesaro-Stolz)}
\\&=\lim_{n\to \infty}\left(\frac{b_{n}^{1+1/k}-b_{n+1}^{1+1/k}}{b_{n+1}^{1+1/k}b_{n}^{1+1/k}}\right)
\\&=\lim_{n\to \infty}\left(\frac{b_{n}^{1+1/k}(1-(1+b_{n}^{1+1/k})^{-1-1/k})}{b_{n+1}^{1+1/k}b_{n}^{1+1/k}}\right)
\\&=\lim_{n\to \infty}\left(\frac{b_{n}^{1+1/k}\left[(1+1/k)b_{n}^{1+1/k}+o(b_{n}^{1+1/k})\right]}{b_{n+1}^{1+1/k}b_{n}^{1+1/k}}\right)
\\&=1+\frac{1}{k}.
\end{aligned}
Thus
$$
\lim_{n\to \infty}\frac{a_n^{k+1}}{n^k}=\left(1+\frac{1}{k}\right)^k
$$
As Christophe pointed out, we can set $\;\theta=\arccos a_1\in[0,\pi]\;.$
First of all, we will prove by induction that
$a_n=\cos\left(\dfrac\theta{2^{n-1}}\!\right)\;\;$ for any $\;n\in\mathbb N\,.\quad\color{blue}{(*)}$
Base case :
if $\;n=1\,,\;$ we get that $\;a_1=\cos\theta=\cos\left(\dfrac\theta{2^0}\!\right)\;,$
hence $\,(*)\,$ is true for $\,n=1\,.$
Induction step :
let $\,k\in\Bbb N\,$ be given and suppose $\,(*)\,$ is true for $\,n=k\,.$
$a_{k+1}=\sqrt{\dfrac{1+a_k}2}=\sqrt{\dfrac{1+\cos\big(\frac\theta{2^{k-1}}\big)}2}=\cos\left(\dfrac\theta{2^k}\!\right)\,.$
Thus, $\,(*)\,$ holds for $\,n=k+1\,$ and the proof of the induction step is complete.
By the principle of induction, $\,(*)\,$ is true for all $\,n\in\Bbb N\,.$
Now, we will calculate the limit
$\lim\limits_{n\to\infty}4^n\big(1-a_n\big)=\lim\limits_{n\to\infty}4^n\left[1-\cos\left(\dfrac\theta{2^{n-1}}\!\right)\right]=$
$=\lim\limits_{n\to\infty}4^n\left(\dfrac\theta{2^{n-1}}\!\right)^{\!2}\left[\dfrac{1-\cos\left(\frac\theta{2^{n-1}}\!\right)}{\left(\frac\theta{2^{n-1}}\!\right)^2}\right]=$
$=\lim\limits_{n\to\infty}4\theta^2\left[\dfrac{1-\cos\left(\frac\theta{2^{n-1}}\!\right)}{\left(\frac\theta{2^{n-1}}\!\right)^2}\right]=4\theta^2\cdot\dfrac12=2\theta^2=$
$=2\big(\!\arccos a_1\!\big)^2\,.$
Best Answer
$\frac{a_n}{a_{n+1}}-1$ must be positive for sufficiently large $n$, so that $(a_n)_{n \ge N}$ is decreasing for some $N$. Also $$ n \cdot\left(\frac{a_n}{a_{n+1}}-1\right) = \frac{n}{\ln n}\cdot \ln n \cdot\left(\frac{a_n}{a_{n+1}}-1\right) \ge \frac{n}{\ln n} \frac 12 \lambda > 2 $$ for sufficiently large $n$, so that $\sum_{n=1}^\infty a_n$ is convergent by Raabe's test.
Finally use that
which is proven e.g. here: If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}{n a_n} = 0$