<s>[INST] Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$. 
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
 \infty} (a_n-3)6^n$.

First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the form $0 \cdot \infty$.
Moreover, we obtained a similar result here, which is related to the form $\lim\limits_{n \to \infty} (a_n-3)9^n$.
How should I proceed?
 [/INST] In this answer we consider the sequence $b_n=|3-a_n|6^n$, where $a_{n+1}=\sqrt{6+a_n}$ with initiial condition $a_0\geq0$. We show that $b_n$ is

*

*bounded and  monotone increasing if $0\leq a_0<3$ (see \eqref{three}).

*bounded  and monotone decreasing when $a_0>3$ (see \eqref{twop}).

As a consequence,


*$b_n(a_0)$ (dependence of initial condition)  is convergent for all $a_0\geq0$:
$$b_\infty(a_0)=|3-a_0|\prod^\infty_{n=1}\Big(1-\frac{|3-a_n|}{6}\Big)^{-1}$$

*When $0\leq a_0<3$, $(a_n-3)6^n=-b_n$ converges to a negative number; when $a_0>3$, $(a_n-3)6^n=b_n$ converges to a positive number.

I don't think the limit $b_\infty(a_0)=\lim_nb_n(a_0)$ can be expressed in terms of elementary functions. A numerical estimate (with double precision arithmetic) gives $b_\infty(0)\approx3.3657$.

The map $f:[0,\infty)\rightarrow[0,\infty)$ defined by
$$ f(x)=\sqrt{6+x}$$
is a contraction: First notice that $f(x)\geq\sqrt{6}$ for all $x\in[0,\infty$, and
\begin{align}
f(x)-f(y)=\frac{y-x}{\sqrt{6+x}+\sqrt{6+y}}\tag{1}\label{one}
\end{align}
Then,
$$|f(x)-f(y)|\leq\frac{1}{2\sqrt{6}}|x-y|
$$
and so,  for any $a_0\geq0$,  $a_{n+1}=f(a_n)$ converges to a fixed  point $x=f(x)$ with $x\geq0$, which in this case is  $x=3$.

Further analysis of the map $f$ shows that

*

*If $3<a_0$, then $3<a_n<a_{n-1}$ for all $n\geq1$

*If $0\leq a_0<3$, then $a_{n-1}<a_n<3$ for all $n\geq1$.


We now study the speed of convergence of the sequence $a_n$. Let $d_n=|3-a_n|=|f(3)-f(a_{n-1})|$. It follows from \eqref{one} that
$$d_n=\frac{d_{n-1}}{3+a_n}$$
If $0\leq a_0<3$, then
\begin{align}
\frac{1}{6}d_{n-1}<d_n=\frac{d_{n-1}}{3+a_n}=\frac{d_{n-1}}{6-d_n}\tag{2}\label{two}
\end{align}
As $a_n\xrightarrow{n\rightarrow\infty}3$, we have that
$\frac{d_{n+1}}{d_n}\xrightarrow{n\rightarrow\infty}\frac{1}{6}$;
consequently $\sqrt[n]{d_n}\xrightarrow{n\rightarrow\infty}\frac{1}{6}$.
This alone however, does not provide convergence of $b_n=(6\sqrt[n]{d_n})^n$.

Now we prove that the sequence $b_n$ indeed converges. From \eqref{two} we have that
\begin{align}
b_{n-1}=6^{n-1}d_{n-1}< 6^nd_n=b_n=\frac{6^{n-1}d_{n-1}}{1-\tfrac{d_n}{6}}=\frac{b_{n-1}}{1-\tfrac{d_n}{6}}\tag{3}\label{three}
\end{align}
and
\begin{align}
\frac{d_{n-1}}{6}<d_n\leq\frac{d_{n-1}}{3+a_0}\leq \frac{d_0}{(3+a_0)^{n-1}}\tag{4}\label{four}
\end{align}
Putting this together, we obtain
\begin{align}
b_n=b_0\prod^n_{k=1}\frac{b_k}{b_{k-1}}=b_0\prod^n_{k=1}\Big(1-\tfrac{d_k}{6}\Big)^{-1}\tag{5}\label{five}
\end{align}
From \eqref{four} we have that  $\sum^\infty_{k=1}d_k<\infty$; from
$$\prod^n_{k=1}\Big(1-\tfrac{d_k}{6}\Big)\leq \exp\Big(-\frac{1}{6}\sum^n_{k=1}d_k\Big)$$
we conclude that $b_n$ converges, and that
\begin{align}
b_n\xrightarrow{n\rightarrow\infty} b_0\prod^\infty_{k=1}\big(1-\tfrac{d_k}{6}\big)^{-1}\tag{6}\label{6}
\end{align}

Remarks:

*

*An interesting observation is that when $a_0>3$, convergence of $b_n$ is much easier to check, for in this case,  the inequalities in \eqref{two} are reverse, that is
\begin{align}
\frac{d_{n-1}}{6+d_n}=\frac{d_{n-1}}{3+a_n}=d_n<\frac{1}{6}d_{n-1}\tag{2'}\label{twop}
\end{align}
This implies  $ b_n<b_{n-1}$ and convergence follows immediately. This can be explained in part to the fact that $f'(x)=\big(2\sqrt{6+x}\big)^{-1}$ is a decreasing function, and so $f'(y)<f'(x)$ whenever $x<3<y$. So convergence tens to be faster to the right of $3$.

*Notice that the limit of the  sequence $b_n$ depends on the initial condition $a_0$. For $a_0=0$, simple numerical implementation gives $\lim_nb_n\approx 3.3657$; for $a_0=6$, $\lim_nb_n\approx 2.7426$
</s>