Square Root Infinite Series #1

If $x$ is $\sqrt{12+\sqrt{12 + \sqrt{12 + \ldots}}}$ , then $x^2 = 12 + \sqrt{12+\sqrt{12 + \sqrt{12 + \ldots}}}$ .

Since $x = \sqrt{12+\sqrt{12 + \sqrt{12 + \ldots}}}$ , $x^2 = 12 + x$ .

We move all terms to the left side: $x^2 - x - 12 = 0$ and factor: $(x + 3)(x - 4)$ .

Solutions: $x = -3$ or $4$ .

Since $\sqrt{12+\sqrt{12 + \sqrt{12 + \ldots}}}$ is definitely going to be positive, $x = 4$ .

Defining the $n$ th term of the series as $a_n = \underbrace{\sqrt{12+ \sqrt{12+ \sqrt{12+ \cdots \sqrt{12}}}}}_{\text{Number of 12's }=n}$ . Then by the ratio test , $\displaystyle \lim_{n \to \infty} \frac {a_{n+1}}{a_n} < 1$ , hence the series converges. Then we can assume:

$\begin{aligned} \blue x & = \blue{\sqrt{12+ \sqrt{12+ \sqrt{12+ \cdots}}}} & \small \blue{\text{Squaring both sides}} \\ x^2 & = 12 + \blue{\sqrt{12+ \sqrt{12+ \sqrt{12+ \cdots}}}} \\ x^2 & = 12 + \blue x \\ x^2 - x - 12 & = 0 \\ (x-4)(x+3) & = 0 & \small \blue{\text{Since }x > 0} \\ \implies x & = \boxed 4 \end{aligned}$

$Solving\space for\space general \space n:$ $let\space \red{x}=\sqrt{n+\red{\sqrt{n+\sqrt{n+\sqrt{n+...}}}}}$ $\Rightarrow \red{x}=\sqrt{n+\red{x}}$ $\Rightarrow \red{x}^2 =n+\red{x}$ $\Rightarrow \red{x}^2-\red{x}-n=0$ $From \space Quadratic\space Formula\space we\space get$ $\red{x}=\dfrac{1{^+_-}\sqrt{1+4n}}{2}$ $\because n\geq 1\therefore \sqrt{1+4n}>2$ $\therefore 1-\sqrt{1+4n}<0 \because \red{x}\geq 0$ $\Rightarrow \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+...}}}}=\dfrac{1+\sqrt{1+4n}}{2}$ $Put \space n=12$ $\Rightarrow \sqrt{12+\sqrt{12+\sqrt{12+...}}}=\dfrac{1+\sqrt{1+4\times 12}}{2}$ $=\dfrac{1+\sqrt{49}}{2}=\dfrac{1+7}{2}=\boxed{4}$