<s>[INST] Find non-zeroes $a$ and $b$ such that $\lim[a^n(u_n - 1)] = b$ where $u_1 = 0$ and $4u_{n + 1} = u_n + \sqrt{6u_n + 3}, \forall n \in \mathbb Z^+$. 
Consider sequence $(u_n)$, defined as $\left\{ \begin{aligned} u_1 &= 0\\ u_{n + 1} &= \dfrac{u_n + \sqrt{6u_n + 3}}{4}, \forall n \in \mathbb Z^+ \end{aligned} \right.$. Knowing that $a$ and $b$ are two real numbers not equal to zero such that $\lim[a^n(u_n - 1)] = b$, calculate the value of $a + b$.

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Why was this question there? WHY WAS THIS IN MY MID-SEMESTER EXAM? (>_<)
I don't need answers, I need to be angered. (This is a joke, please take it with a grain of salt. I do actually need answers to this problem.)
Here are my observations.
According to the WolframAlpha, the general term of the sequence above is $$u_n = \left(1 - \dfrac{1}{2^{n - 1}}\right)\left(1 - \dfrac{2 - \sqrt 3}{2^{n - 1}}\right), \forall n \in \mathbb Z^+$$
If we define sequences $(v_n)$ and $(w_n)$ as $v_n = 1 - \dfrac{1}{2^{n - 1}}, \forall n \in \mathbb Z^+$ and $w_n = 1 - \dfrac{2 - \sqrt 3}{2^{n - 1}}, \forall n \in \mathbb Z^+$ respectively, then it can be obtained that $u_n = v_nw_n, \forall n \in \mathbb Z^+$ and $$\left\{ \begin{aligned} v_1 &= 0\\ v_{n + 1} &= \dfrac{v_n + 1}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right. \text{ and } \left\{ \begin{aligned} w_1 &= -1 + \sqrt 3\\ w_{n + 1} &= \dfrac{w_n + 1}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right.$$
That means $$\begin{aligned} v_{n + 1}w_{n + 1} = \dfrac{v_nw_n + \sqrt{6v_nw_n + 3}}{4} &\iff \dfrac{(v_n + 1)(w_n + 1)}{4} = \dfrac{v_nw_n + \sqrt{6v_nw_n + 3}}{4}\\ &\iff v_n + w_n + 1 = \sqrt{6v_nw_n + 3}\\ &\iff w_n = (2v_n - 1) + \sqrt 3(1 - v_n), \forall n \in \mathbb Z^+ \end{aligned}$$
Anyhow, about the sequence $(u_n)$ itself, it is strictly increasing bounded. More specifically, we have that $u_n \in [0; 1), \forall n \in \mathbb Z^+$. The same goes for sequences $(v_n)$ and $(w_n)$.
Actually, $v_{n + 1} = \dfrac{v_n + 1}{2}$ and $w_{n + 1} = \dfrac{w_n + 1}{2}$, those look familiar, hmmm~
Of course, I forgot. If we let $v_n = \cos a_n, \forall n \in \mathbb Z^+$ and $w_n = \cos b_n, \forall n \in \mathbb Z^+$, then it can be obtained that $$\left\{ \begin{aligned} a_1 &= \dfrac{\pi}{2}\\ \cos(a_{n + 1}) &= \cos^2\dfrac{a_n}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right. \text{ and } \left\{ \begin{aligned} b_1 &= \arccos(-1 + \sqrt 3)\\ \cos(b_{n + 1}) &= \cos^2\dfrac{b_n}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right.$$
Nevermind, that didn't work as well as I had thought.
What was I doing this entire time? You might be wondering. Well, I'm trying to find the general term of sequence $(u_n)$ without the need of a laptop, since you can't take that into the testing room.
Anyhow, for the second part of the problem, first of all, let $\lim u_n = m$, then we have that $m = \dfrac{m + \sqrt{6m + 3}}{4} \iff m = 1$. Again, the same goes for sequences $(v_n)$ and $(w_n)$. Futhermore, $$\begin{aligned} \left\{ \begin{aligned} 2^n(v_n - 1) &= -2\\ 2^n(w_n - 1) &= 2\sqrt 3 - 4 \end{aligned} \right. &\iff \left\{ \begin{aligned} 2^n(v_n + w_n - 2) &= 2\sqrt{3} - 6\\ 4^n(v_n - 1)(w_n - 1) &= 8 - 4\sqrt{3} \end{aligned} \right.\\ &\implies 4^n\left[u_n - \left(\dfrac{2\sqrt{3} - 6}{2^n} + 2\right) + 1\right] = 8 - 4\sqrt{3}\\ &\iff 4^n(u_n - 1) = (8 - 4\sqrt{3}) - 2^n(6 - 2\sqrt{3})\\ &\iff 2^n(u_n - 1) = \dfrac{(8 - 4\sqrt{3})}{2^n} - (2\sqrt{3} - 6), \forall n \in \mathbb Z^+\\ &\implies \lim[2^n(u_n - 1)] = 2\sqrt{3} - 6 \end{aligned}$$
In conclusion, $a + b = 2 + (2\sqrt{3} - 6) = 2\sqrt{3} - 4$.
My question is more focused on the first part of the problem, on how the general term of sequence $(u_n)$. As always, thanks for reading (and even more if you could help~) By the way, the options were $-1, \sqrt 3 - 1, 2\sqrt 3 - 4$ and $4 - 2\sqrt 2$.
 [/INST] Well, there may be something "lost in translation", indeed: if all those questions were multiple choice, which were the choices for $a+b$?
I'm asking because $u_n$ is converging to $1$ quite rapidly, and since
$$a=\lim_{n\to\infty}\frac{u_n-1}{u_{n+1}-1},$$ computing that ratio for a few $n$ (up to $n=10$, say) with a pocket calculator would suggest $a=2$, and calculating $b_n=2^n\,(u_n-1)$ for $n=10$ might be sufficient to identify $a+b$ among the given choices.
If you really want to discover the closed form, it's hardly a good idea to start from WolframAlpha's answer. Instead, let's make that square root rational. And since the limit of $u_n$ is $1$, and thus, the limit of $6\,u_n+3$ is $9$, let
$$6\,u_n+3=9\,v^2_n.$$ This gives
$$u_n=\frac{3\,v^2_n-1}2,$$ and
$$\frac{3\,v^2_{n+1}-1}2=\frac{\frac{3\,v^2_n-1}2+3\,v_n}4$$ simplifies to
$$v^2_{n+1}=\frac{v^2_n+2\,v_n+1}4=\left(\frac{v_n+1}2\right)^2,$$
i.e.
$$v_{n+1}=\frac{v_n+1}2.$$ So
$$v_{n+1}-1=\frac{v_n-1}2,$$ meaning
$$v_n-1=(1/2)^{n-1}(v_1-1)=(1/2)^{n-1}\left(\sqrt{1/3}-1\right).$$ This gives another closed form for $u_n$, but proving this to be identical with WolframAlpha's answer may be left as an exercise to the reader. ;-)
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