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How to find the value of $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$

I tried to solve it and found a relation that if I assume the given expression to be something say $x$, then following result holds true. $$\boxed{x = 3 + \sqrt{3x}}\hspace{4cm}\bf ...(1.)$$

Can be proved as, $$x = 3 + \sqrt{3(3 + \sqrt{3x})}$$ $$x = 3 + \sqrt{3^2 + \sqrt{3^3x}}$$ $$x = 3 + \sqrt{3^2 + \sqrt{3^3(3 + \sqrt{3x})}}$$ $$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7x})}}$$ $$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7(3 + \sqrt{3x})})}}$$ $$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{15}x}}}}$$

If I continue this process repeatedly, $$x= 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^{8} + \sqrt{3^{16} +\sqrt{...}}}}}$$

But how can I make sure that $(1.)$ is absolutely correct? Also how can I make sure that "If I continue this process repeatedly" I would get the same expression?

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    The important question here is how do you define this infinite expression, usually it is a limit of finite expressions. You just need to prove that the limit exists and formulate $x = 3 + \sqrt{3x}$ in the sense of limits. – ajr Apr 06 '22 at 08:29

3 Answers3

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Let $$f(x)=x+\sqrt{x^2+\sqrt{x^4+\sqrt{ x^8+\cdots}}},$$ which is to be understood as $\lim_{n\to\infty}f_n(x)$, where $$f_n(x)=x+\sqrt{x^2+\sqrt{x^4+\sqrt{ x^8+\cdots\sqrt{\cdots\sqrt{x^{2^n}}}}}},$$ or recursively (and thus finally in a sufficiently formal way), $$\tag1f_0(x)=x,\qquad f_{n+1}(x)=x+\sqrt{f_n(x^2)}.$$ We want to find the limit (if it exists) $f(3)$. Note that instead of your equation, $(1)$ rather gives us $$\tag2 {f(x)=x+\sqrt{f(x^2)}}.$$ Your guess seems to have ignored the way the exponents of $x$ grow under the deeper radicals.

The above looks like we need to first find $f$ evaluated at larger arguments (hence with seemingly "worse" convergence), but we can easily get out of that trap:

For $x>0$, let $g_n(x)=\frac1xf_n(x)$. Then $(1)$ becomes $$ g_0(x)=1,\qquad g_{n+1}(x)=1+\frac1x\sqrt{f_n(x^2)}=1+\sqrt{\frac1{x^2}f_n(x^2)}=1+\sqrt{g_n(x^2)}.$$

In particular, we find by induction that all $g_n$ are constant, $g_n(x)=G_n$. What about convergence of this sequence of constants $$\tag3 G_0=1,\qquad G_{n+1}=1+\sqrt{G_n}?$$

As $x\mapsto 1+\sqrt x$ is an increasing function on $[0,\infty)$ and as $G_1=2>1=G_0$, we see that $\{G_n\}_n$ is inceasing. On the other hand, we readily see by induction that $G_n<3$. Hence $G:=\lim_{n\to\infty}G_n$ exists, and certainlky $1<G\le 3$.

From $(2)$, we know that $G$ must be a soution of $$ x=1+\sqrt{x}$$ or,

$$ (x-1)^2=x.$$

This quadratic has solutions $x_{1,2}=\frac{3\pm\sqrt{5}}2$, and as we know $G>1$, we have $G=\frac{3+\sqrt{5}}2$ and ultimately

$$f(3)= 3g(3)=3G=\frac32(3+\sqrt{5}).$$

Hagen von Eitzen
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  • $x = 3 + \sqrt{3x}$ also gives the same result $x = \dfrac{3}{2}(3 + \sqrt{5})$ ._. –  Apr 06 '22 at 09:08
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    @user983206 Intriguing. Then perhaps my judgement was too quick and your method works fine after all. However, I think that it gets into trouble because it has worse problems with convergence in the first place. In fact that is precisely the doubt about "If I continue this process repeatedly" that made you ask this question in the first place. – Hagen von Eitzen Apr 06 '22 at 09:13
  • You maintain the suspense up to the end, Mr Hitchcock ! Cheers – Claude Leibovici Apr 06 '22 at 12:58
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I think that OP's guess was okay, here's a different approach.

Let $a_1 = 3$ and let $a_{n+1} = 3 + \sqrt{3a_n}$ for $n\geq 1$. Then the first few elements are \begin{align*} a_1 &= 3,\\ a_2 &= 3+ \sqrt{3^2},\\ a_3 &= 3+\sqrt{3^2 + \sqrt{3^4}},\\ a_4 &= 3+ \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8}}}, \end{align*} so it is reasonable to define \begin{align*} 3+ \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8+\ldots}}} := \lim\limits_{n\to\infty} a_n. \end{align*} We first need to make sure that the limit exists. We will use the theorem saying that any monotone and bounded sequence is convergent, it is often useful for recursive sequences such as $a_n$.

Monotonicity: we will prove it by induction. First note that $a_2 \geq a_1$. Now assume that for some $n\geq 1$ we have $a_{n+1}\geq a_n$. Then \begin{align*} a_{n+2} - a_{n+1} = 3 + \sqrt{3a_{n+1}} - (3 + \sqrt{3a_{n}}) = \sqrt{3}(\sqrt{a_{n+1}} - \sqrt{a_n}) \geq 0, \end{align*} where the last inequality follows from the induction hypothesis and the fact that the square root is increasing. Thus, by induction we get that $a_{n+1}\geq a_n$ for all $n\geq 1$, so the sequence is increasing.

Boundedness: obviously the sequence is bounded from below by 0. To find a reasonable upper bound let us first compute what we think the limit should be (if a sequence is increasing, then its limit is also its upper bound). Note that \begin{align*} \lim\limits_{n\to \infty} a_{n} = \lim\limits_{n\to \infty} a_{n+1} = 3 +3\lim\limits_{n\to\infty}\sqrt{a_n}, \end{align*} so if we let $x = \lim\limits_{n\to \infty} a_{n}$, we get that $x = \frac 32(3 + \sqrt{5}).$ To make the computations easier, we will show that $a_n \leq 27$ by induction ($27\geq x$ and it works nicely if we plug it into $\sqrt{3\cdot a_n}$). It is of course true for $a_1$. If $a_n\leq 27$ , then \begin{align*} a_{n+1} = 3 + \sqrt{3a_n} \leq 3 + \sqrt{3\cdot 27} = 3+9 = 12 \leq 27. \end{align*} Thus we get that $a_n$ is bounded and increases, so it has a finite limit. By the above computation, the limit is equal to $x$.

ajr
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If we have $$a_{n+1}=x \sqrt{x\,a_n} \qquad \text{with}\qquad a_0=x$$

$$a_n=x^{b_n} \implies x^{b_{n+1}}=x \sqrt{x^{b_n+1}} \implies b_{n+1}=\frac 12 b_n+\frac 32$$ $$b_n=c_n+3\implies c_{n+1}=\frac 12 c_n$$

Finishing and back to $a_n$ $$a_n=x^{3-2^{1-n}}\quad \to \quad x^3$$

Claude Leibovici
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