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?
 A: Here's a proof that $b_n:=(3-a_n)6^n$ converges to a finite limit. This is merely a proof of existence of a finite limit (which I don't think is trivial).
It is shown here that $a_n$ is strictly increasing and has limit $3$.
Note that
$$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}$$
hence
$$(3-a_{n+1})6^{n+1} = \frac{6}{3+\sqrt{6+a_n}} (3-a_n)6^n$$
thus
$$\frac{b_{n+1}}{b_n} = \frac{6}{3+\sqrt{6+a_n}}>1$$

Let us show that $\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.
From $$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}\leq \frac 13 (3-a_n)$$
one readily finds that $3-a_n = O(\frac 1{3^n})$, hence $a_n = 3 + O(\frac 1{3^n})$, thus
$$\ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big) 
= \ln\Big(\frac{6}{3+\sqrt{9+O(\frac 1{3^n})}}\Big) 
= \ln(1+O(\frac 1{3^n})) 
= O(\frac 1{3^n})$$
Hence$\sum_n \ln\Big(\frac{6}{3+\sqrt{6+a_n}}\Big)$ converges.

Thus $\sum_n \log(\frac{b_{n+1}}{b_n})$ converges, hence the sequence $\log b_n$ converges to a finite limit, hence $b_n$ converges as well to some positive real.
Therefore, the sequence $(a_n-3)6^n$ converges to a negative real number.
A: 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 $a_0=6$, $\lim_nb_n\approx 2.7426$
A: The function, $a_1 \mapsto \Phi(a_1) = \lim_{n\to \infty} 6^n (a_n-3)$  is a well-known object in complex dynamics and known as a linearization map. It solves a conjugation problem, which was first introduced by Ernst Schröder in 1871 to study iterations of rational functions. You may find an account e.g. in Carleson and Gamelin: Complex Dynamics. In chap II.2 we have the following:
Theorem 2.1: Suppose an analytic function $f$ has an attractive fixed point $p$ with multiplier $\lambda=f'(p)$ satisfying $0<|\lambda|<1$. Then there is a conformal map $\zeta=\phi(z)$ (unique up to scaling) of a neighborhood of $p$ onto a neighborhood of $0$ which conjugates $f(z)$ to the linear function $g(\zeta)=\lambda \zeta$.
The proof goes through the construction of the limit you mention. To be more explicit and using your example: Let $D={\Bbb C}\setminus (-\infty;-6]$ be the slit complex plane. Then $f(z)=\sqrt{z+6}$ defines a conformal map of $D$ into the right half plane ${\Bbb H}_+\subset D$. Thus, for all $z\in D$: $$|f(z)-3| = \left|\frac{z-3}{\sqrt{z+6}+3}\right| \leq \frac13 |z-3|$$
Thus we have convergence of $f^n(z)$ to the fixed point $p=3$, uniformly on compact subsets of $D$.
We also have $\lambda=f'(3) = \frac16$.
The proof of the theorem is then to show that the sequence of maps ($n\geq 0$):
$$ \phi_n(z) = \lambda^{-n} (f^n(z)-p) = 6^n (f^n(z)-3)$$
converges uniformly on a neighborhood of $p$.
Pick $z_0=z\in D$ and let $z_n=f^n(z_0)$ (I prefer starting at index 0 so that we iterate $f$ as many times as the linear map) we have:
$$ \phi_{n+1}(z) = 6^n (z_n-3) \frac{6}{\sqrt{z_n+6}+3} =
\phi_n(z) \left(1 + \frac{3-\sqrt{z_n+6}}{\sqrt{z_n+6}+3}\right) =
\phi_n(z) \left( 1 + \epsilon(z_n)\right)
$$
where  $$\epsilon(z_n) = \frac{3-z_n}{\left( \sqrt{6+z_n} + 3\right)^2}. $$
One has $|\epsilon(z_n)| \leq \frac19 |z_n-3| \leq \frac{1}{3^{n+2}} |z-3|$
thus going to zero exponentially fast and uniformly on compact subsets in $D$. It follows that the limit $$\phi(z) = \lim_n \phi_n(z)=(z-3)\prod_{n\geq 1} (1+\epsilon(z_n))$$ exists and is holomorphic in $D$.
The function verifies $\phi'(3)=1$ whereas your function has an extra factor $6$ since your index starts with $n=1$ rather than zero. So your function is really $\Phi(a_1)=6\phi(a_1)$. The formula is numerically precise and efficient: $6\phi(\sqrt{6})=-3.36565753974384...$ in agreement with other results posted.
To see that $\phi$ actually yields a solution to the Theorem, note that by construction we have $\lambda^{-1} \phi_n\circ f = \phi_{n+1}$. By uniform convergence we may take limits to get $\lambda^{-1} \phi\circ f = \phi$ or $\phi \circ f = \lambda \phi = g\circ \phi$, with $g(\zeta)=\lambda \zeta$ being the linear map.
A Taylor expansion of $\phi$ may be obtained from the functional equation $6\phi(\sqrt{z+6}) =  \phi(z)$ by expanding at $z_0=3$ on both sides and identifying coefficients. A far more elegant way to do the computations was suggested in comments by @VarunVejalla: Solving $w=\sqrt{z+6}$ gives $z=w^2-6$, i.e. the relation $6\phi(w)= \phi(w^2-6)$. To simplify notation set $w=3+u$. Then $\psi(u)=\phi(3+u)$
is to be expanded around 0 and verifies $6\psi(u)=\phi((3+u)^2-6) = \psi(6u+u^2)$. Inserting $\psi(u)=u + \sum_{n\geq 2} a_n u^n$ yields the identity between Taylor series:
$$ 6u + \sum_{n\geq 2} 6 a_n u^n = (6u+u^2) + \sum_{n\geq 2} a_n (6u +u^2)^n $$
or
$$ \sum_{n\geq 2} a_n ((6u+u^2)^n-6u^n)  = -u^2 $$
Expanding the binomial and identifying coefficients you may solve recursively to get a formula for $a_n$ which only depends upon the $a_k$ with $n/2 \leq k \leq n$. You get $a_1=1$ and then
$$ a_n = \frac{1}{6-6^n} \sum_{k =\lceil n/2 \rceil}^{n-1}
\pmatrix {k \\ n-k} 6^{2k-n} a_k $$
The first terms:
$$ \psi_4(u) = u -\frac{1}{30}u^2 + \frac{1}{525} u^3 - \frac{181}{1354500} u^4. $$
With 12 terms one gets $6 \psi_{12}(\sqrt{6}-3) =  3.365657539743842...
$ correct to 14 digits.
A: The following solution is a special case of a general method.
The question asks

Let $a_ 1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find
$\lim_{n \to \infty} (a_n-3)6^n$.

With $\,q\,$ as a parameter, define the function
$$ F(x) := \sum_{n=0}^\infty c_n \frac{x^n}{f_n}\;\;
 \text{ where }\;\; f_n:= \prod_{k=1}^n (1-q^k) \tag{1} $$
and where $\,|q|\ne1.\,$ Define the constants
$$ L := q/2, \quad \text{ and } \quad K := L^2-L. \tag{2} $$
Also define
$$ a_n := A\left(\frac{x_0}{q^n}\right) \;\;
\text{ where } \;\; A(x) := L - q\, x\, F(x) \tag{3} $$
and where $\,x_0 = A^{-1}(a_0)\,$ depends only on
$\,q\,$ and $\,a_0.\,$
Tthe value of $\,x_0\,$ computed this way does not
suffer from floating point rounding problems.
The equation
$$ a_{n+1} = \sqrt{K+a_n} \quad \text{ or } \quad a_{n+1}^2 = K+a_n \tag{4} $$
implies that  the coefficients of $\,F(x)\,$ as polynomials in $\,q\,$ satisfy
$$ c_0 = 1, \quad c_{n+1} = \sum_{k=0}^n c_{n-k}\,c_k\,
 \frac{f_n}{f_k f_{n-k}}. \tag{5} $$
Note that
$$ F(0) = 1,\; A(0) = L \; \text{ and }\;
 a_n\to L. \tag{6} $$
The approach of $\,a_n\,$ to the limit $\,L\,$ is given by
$$ L - a_n = q \, \frac{x_0}{q^n}
 F\left(\frac{x_0}{q^n} \right)
\approx q\frac{x_0}{q^n}. \tag{7} $$
This implies
$$ \lim_{n\to\infty}(a_n-L)\,q^n = -q\,x_0. \tag{8} $$
In the case in the question,
$$ q = 6,\; L = 3,\; a_0 = 0,\;
 x_0 \approx 0.56094292329064. \tag{9} $$
In the case from Art of Problem Solving Online,
$$ q = 4,\; L = 2,\; a_0 = 0,\;
 x_0 = \frac{\pi^2}{16},\; F(x) = \frac{\sin(\sqrt{x})^2}x. \tag{10} $$

NOTE: About my method.
Some of it is based on Koenigs function but my version is more constructive.
If we have a function $\,T(x)\,$ and define a sequence by
$\,a_{n+1} = T(a_n)\,$ where $\,a_n\to 0\,$
such that $\, a_n \approx c/q^n,\,$ then use the Ansatz
$\, a_n = F(x/q^n)\,$ for some function $\,F\,$ with a power
series expansion in $\,x\,$ with coefficients that depend on $\,q.\,$
The coefficients are uniquely determined by the function $\,T.$
The mode of convergence determines the proper Ansatz. For
example, $\,T(x) := x-x^2\,$ requires a different Ansatz.
NOTE: If $\,q=1\,$ then the convergence $\,a_n\to \frac14\,$ is
much slower and the above analysis does not hold. Instead the one
for $\,T(x) := x-x^2\,$ is needed. Something similar if $\,q=-1\,$
and other roots of unity.
NOTE: I was going to mention chapter 8.3 of Asymptotic Methods in Analysis
by de Bruijn which I am familiar with but didn't have the exact page number.
A: Ths approach shows that the results in question emerge quite naturally from the standard fixed-points-stability analysis of the evolution equation. Also the dependence of the limit of $b_n = (a_n -3)6^n$  from the inital value $a_1$ is discussed.
Following the common procedure to analyse the evolution equation
$$a_{n+1} = \sqrt{6+a_{n}}\tag{1}$$
we fist look for fixed points. These follow from the equation $0 = a^2-a-6$ to be $a = 3$ and $a=-2$.
Ruling out the negative solution we study the stabilty letting
$$a_{n} = 3 +\delta_{n}\tag{2}$$
where $|\delta_{n}| << 3$.
Inserting into $(1)$ gives
$$\begin{align}
& 3 +\delta_{n+1} = \sqrt{6+3+\delta_{n}}=\sqrt{9+\delta_{n}}\\&=3\left( \sqrt{1+\frac{\delta_{n}}{9}}\right)\simeq 3\left( 1+\frac{1}{2}\frac{\delta_{n}}{9}\right)=3 +\frac{1}{6} \delta_{n}\end{align} $$
whence follows
$$\delta_{n+1} \simeq \frac{1}{6} \delta_{n}\tag{3}$$
The solution $\delta_{n}=\frac{c}{6^n}$ of $(3)$ with some constant $c$ shows that the fixed point $a_{\infty}=3$ is stable and from $(2)$ we find
$$a_{n} \simeq 3 + \frac{c}{6^n}\tag{4}$$
This in turn means that
$$b_{n} := \left(a_{n} - 3 \right)6^n \simeq c\tag{5}$$
So that the limit in question is the constant $c$. For $0 \lt a_{1} \lt 3$ the constant $c$ is negative, for $a_{1} \gt 3$ it is positive. In the actual case we have $a_{1} = \sqrt{6} \lt 3$, and the limit is numerically
$$b_{\infty} \simeq -3.36571\tag{6}$$.
Limit as a function of the initial value
Here is a plot showing the dependence of the limit $b_{\infty}$ on the initial value $a_{1}$ for $-6 \le a_{1} <10$

Some special values of the limit are: $b_{\infty}(-6) = -121.164$, $b_{\infty}(0) = -20.1939$, and $b_{\infty}(3) = 0$
The limit function is well approximated by
$$b_{\infty}(a_1)=45.1116 (a_1+6)^{0.44966}-121.164\tag{7}$$
A: Found a possible numerical approach compatible with Gabriel Romon above. Calculate $a_n$ for $n=0,1,2,...,10$.
Define $q_n=(3-a_n)6^{n}$.
Then, $\ln(1-\frac{q_n}{q_{n+1}})$ can be fitted with a straight line as a function of $n$  having $r^2=0.9983$.
If the argument of the natural logarithm here is taken to be the ratio of the first derivative to the function itself, this implies $q(n)$ is a double exponential.
To Wit:
$$q_n=(3-a_n)6^n \approx q(n)= c_1e^{(\frac{e^b}{m}e^{mn})}$$
where m and b are the slope and intercept respectively for the aforementioned line, and $c_1$ is a constant.
In my modeling, m=-1.8368, b = -0.4174.
This suggest $c_1$ is the limit in question and fitting $q(n)$ against the $q_n$ will yield $c_1$, approximating the desired result, however so far my attempts give a value of about 4.27.
A: With $b_n:=(a_n-3)6^{n}$ we rewrite
$$6^{-n-1}b_{n+1}+3=\sqrt{6+6^{-n}b_n+3}$$
and
$$b_{n+1}=6^{n+1}\left(\sqrt{9+6^{-n}b_n}-3\right)=\frac{6b_n}{\sqrt{9+6^{-n}b_n}+3},$$
with $b_1=6(\sqrt 6-3)$. This proves that if $b_n$ converges, it converges to a negative value. As the denominator is asymptotic to $a+6^{-n}c$, it should be possible to find an upper bound and prove convergence.
