Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$

Question

Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$ My approach: Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$ so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$ So, the answer is $\boxed{4}$.

---

Is correct my solution? Can you show other ways for to solve this problem? Can you suggest me any textbooks with similar problems? Thank you so much!

Answer 1

You need to show the sequence $$ x_n = \sqrt{12 + \sqrt{12 + \cdots \text{n times} }} $$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps. It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$. As answered by @user2661923 and @QC_QAOA the sequence $x_n$ is bounded. Here is an easy way of showing $x_n$ is strictly increasing. Observe that $x_1 < x_2$ and we assume that $x_{n-1} < x_n$. Then $$x_{n+1} = \sqrt{12 +x_n} > \sqrt{12 + x_{n-1}} = x_n$$ which completes our inductive argument and we conclude that $x_n$ converges.

Observing that $x_n > 0$ for any $n$ we can conclude that $x \ge0$ by limit theorems. So we must have $x = 4$

Answer 2

Edit
First see the comment of NinadMunshi immediately following this answer. I have edited the answer accordingly.

---

Continuing Infinity_hunter's answer, and letting

$$x_n~ \text{denote}~ \sqrt{12 + \sqrt{12 + \cdots \text{n times} }} $$

To show convergence, all that is necessary is to show that

• the sequence is bounded
• the sequence is strictly increasing.

Clearly, $0 < x_1 < 4.$ Assume that $0 < x_n < 4.$ Then

$$x_{(n+1)} = \sqrt{12 + x_n} < \sqrt{12 + 4} < 4$$

and

$$x_{(n+1)} = \sqrt{12 + x_n} > \sqrt{12 + 0} > 0.$$

Therefore,

$$0 < x_n < 4 \implies 0 < x_{(n+1)} < 4.$$

To show that the sequence is (therefore) strictly increasing:

Since

$$ 0 < x_n < 4$$

and

$$[(x_n)^2 - x_n - 12] = (x_n - 4)(x_n + 3)$$

I conclude that

$$[(x_n)^2 - x_n - 12] < 0 \implies (x_n)^2 < x_n + 12.$$

However, by the definition of the sequence

$$\left[x_{(n+1)}\right]^2 = x_n + 12.$$

Therefore

$$\left[x_{(n+1)}\right]^2 > (x_n)^2.$$

Therefore, since each element in the sequence is positive,

$$x_{(n+1)} > (x_n).$$

Therefore, the sequence is strictly increasing.

Answer 3

Let $a_1=\sqrt{12}$ and $a_n=\sqrt{12+a_{n-1}}$. We must show that $a_n\to 4$. In order to do this, it is sufficient to show that

$1)\ \lim_{n\to\infty}a_n\text{ exists}$

$2)\ \text{This limit is }4$

You have basically skipped step $1)$ and gone straight to step $2)$. To fill in the first step, first note that $0<a_n<4$ for all $n$. We proceed by induction:

$$0<a_1=\sqrt{12}<\sqrt{16}=4$$

$$0<\sqrt{12}\leq a_n=\sqrt{12+a_{n-1}}<\sqrt{12+4}=\sqrt{16}=4$$

Second, we may now show that $a_n$ is increasing. This is obvious as

$$\frac{a_{n+1}}{a_n}=\frac{\sqrt{12+a_n}}{a_n}=\sqrt{\frac{12}{a_n^2}+\frac{1}{a_n}}$$

Since $a_n<4$, this becomes

$$\sqrt{\frac{12}{a_n}^2+\frac{1}{a_n}}>\sqrt{\frac{12}{16}+\frac{1}{4}}=\sqrt{\frac{3}{4}+\frac{1}{4}}=\sqrt{1}=1$$

Thus, $a_{n+1}>a_n$. Since we have already established that $a_n<4$ we may conclude that

$$\lim_{n\to\infty}a_n=L$$

exists. This concludes step $1)$. As you have already proved step $2)$, we may conclude that

$$\lim_{n\to\infty}a_n=4$$

The reason we may discard the $-3$ answer is that $0<a_n<4$ so $L\in[0,4]$.

---

EDIT: Edited as OP requested more resources for these types of infinite radicals. The best place to start (at least in my mind) is Herschfeld's Convergence Theorem. The theorem states that

$$\lim_{k\to\infty} \left[x_0+(x_1+(x_2+(\cdots +(x_k)^p)^p)^p)^p\right]$$

exists if and only if $(x_n)^{p^n}$ is bounded. We see that at each step, $x_n=12$. Since $12^{2^{-n}}$ is bounded, we may conclude that the series converges.

Answer 4

Your solution is correct. The other answers have addressed formality and rigour as a concern, but from the sound of your comments, you're more interested in the kind of problem more than the right answer, so I hope the following helps.

This kind of problem is an infinitely nested radical. As far as I know, there's no other way as straightforward as recognizing the infinitely nested square root is just a quadratic in disguise. In fact, an infinitely nested cube root would be a cubic in disguise.

Solution of Polynomial Equations with Nested Radicals on Arxiv actually derives a generic nested radical as a solution to a generalized quadratic equation, and dives deeper into different kinds of polynomials.

On Infinitely Nested Radicals is a less crazier article from a magazine that discusses their convergence and what type of numbers could be represented by them.

A Chronology of Continued Square Roots and Other Continued Compositions [...] on Arxiv is exactly what it sounds like. At a glance it's written pretty well.

Answer 5

Method 1 - contraction mapping theorem.

Let $g : [0,\infty) \to [0,\infty)$ be the map $x \mapsto \sqrt{x+12}$. For any $x, y \in [0,\infty)$, we have

$$\begin{align} & g(x) - g(y) = \sqrt{x+12} - \sqrt{y+12} = \frac{x - y}{\sqrt{x+12} + \sqrt{y+12}}\\ \implies & |g(x)-g(y)| \le \frac{|x-y|}{2\sqrt{12}} \end{align} $$ This means $g$ is a contraction mapping over $[0,\infty)$.

By Contraction mapping theorem, $g(x)$ have a unique fixed point over $[0,\infty)$. Furthermore, if one pick any $z \in [0,\infty)$ and construct a sequence $z_n$ by $$z_n = \begin{cases}z, &n = 0\\g(z) = \sqrt{z+12}, & n > 0\end{cases}$$ $z_n$ will converges to that fixed point.

Since $g(4) = 4$, that unique fixed point is $4$. By setting $z$ to $0$, we find

$$\begin{array}{rcl} z_1 &=& \sqrt{12},\\ z_2 &=& \sqrt{12+\sqrt{12}},\\ z_3 &=& \sqrt{12+\sqrt{12+\sqrt{12}}}\\ &\vdots& \end{array}$$ converges to $4$.

Method 2 - explicit bound.

Define $z_n$ as in method $1$ and let $y_n = 4 - z_n$ for $n \ge 1$. Notice $$y_{n+1} = 4 - z_{n+1} = 4 - \sqrt{12 + z_n} = 4 - \sqrt{16-y_n} = \frac{y_n}{4 + \sqrt{16-y_n}}\tag{*1} $$ $y_{n+1}$ has same sign as $y_n$. Since $y_1 = 4 - \sqrt{12}> 0$, all $y_n$ are positive.
Notice $\sqrt{16-y_n} = \sqrt{12+z_n} \ge 12$, $(*1)$ implies

$$0 < y_{n+1} < \frac{y_n}{4+\sqrt{12}}$$

Replace $n$ by $1, 2, \ldots, m-1$ and combine the inequalities, we find for $m \ge 1$,

$$0 < y_m \le \frac{y_1}{(4+\sqrt{12})^{m-1}}\quad \implies\quad 4 - \frac{4}{(4+\sqrt{12})^m} \le z_m < 4$$ By squeezing,

$$\lim_{m\to\infty} z_m = \lim_{m\to\infty} \underbrace{\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}}_{m\text{ times}} = 4$$