<s>[INST] Why is $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = 3$ This is a problem from SASMO Grade 8 (Secondary 2) Sample Questions.
Solve for $x$
$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 3$
Answer: $x=6$
I have tried this on a calculator: the more $x$ we add, the closer we are to $3$. But how can we prove it, and is there a way to figure this out by hand?
 [/INST] Let $x=\sqrt{6 + \sqrt{6 + \sqrt{6 + …}}}$ Next, square both sides: $x^2=6 + \sqrt{6 + \sqrt{6 + …}}$ Then, subtract 6 from each side: $x^2-6=\sqrt{6 + \sqrt{6 + …}}$ Wait a minute..... the right side of the equation is x! Substituting in, we get $x^2-6=x$, and $x^2-x-6=0$. This is a simple quadratic we can solve, yielding the solutions $x=3$ and $x=-2$. But which solution is the real one? Actually, when we squared both sides of the original equation, we introduced an extranneous solution. This extranneous solution is -2, and notice that when we substitute it back into the original equation we get $-2=\sqrt{6-2}$   ->   $-2=2$. This is obviously wrong, but it makes sense that we got this solution from the quadratic because when we square both sides we get $4=4$, which is correct. Therefore $x=\sqrt{6 + \sqrt{6 + \sqrt{6 + …}}}=3$
EDIT
Notice that by solving for x using arithmetic we are assuming that x is a finite number and not positive or negative infinity. We can show this by letting $a_k$ represent a 'nest' of $k$ $\sqrt6$'s (e.g. $a_3=\sqrt{6 + \sqrt{6 + \sqrt{6}}}$) and proving via induction that if $a_n$ is between 0 and 3 then $a_{n+1}$ is also between 0 and 3: notice that the rule $a_{n+1}=\sqrt{6+a_{n}}$ applies to all integers $n>1$. We then apply the following logic:
$0<a_{n}<3$
$6<6+a_{n}<9$
$\sqrt6<\sqrt{6+a_{n}}<3$
$0<a_{n+1}<3$
Since $a_1=\sqrt6$ is obviously between 0 and 3, we know the sequence does not diverge.
However, we are not done here! Just because we know that $a_n$ is finite for all n does not tell us that the sequence converges, e.g. $1-1+1-1+1-...$ does not converge although all partial expressions $1, 1-1, 1-1+1, 1-1+1-1, ...$ are finite. We can show the aforementioned sequence does not behave like this by showing $a_{n+1}>a_n$ for all n. Both sides are greater than 0 so we can square both sides. Using the previously stated recursion $a_{n+1}=\sqrt{6+a_{n}}$ the inequality is equivalent to saying $6+a_{n}>a_{n}^2$, so $0>a_{n}^2-a_n-6$. If we solve the quadratic we see that the inequality is true for all $a_n$ such that $-2<a_n<3$, which is true. Therefore we know the series converges to a finite value.
Q.E.D.
tl; dr: THE SERIES CONVERGES, AND IT INDEED CONVERGES TO 3.
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