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Let $\{a_n\}$ be the sequence defined by:

$a_1 = \sqrt{2}$, $a_{n+1}=\sqrt{2+\sqrt{a_n}}$

Show that $\{a_n \}$ is convergent and find $\lim_{n \to \infty} a_n$

My attempt

I have already shown that $a_n$ is increasing and is upper bounded by $3$ and therefore the sequence is convergent, but I'm having trouble finding the limit of convergence.

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

Then since the subsequence $\{ a_{n+1} \}$ also converges to the same limit we have: $L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{2+\sqrt{a_n}} = \sqrt{2+\sqrt{L}}$ (since $\sqrt{2+\sqrt{x}}$ is a continuous function for $x \geq 0$). Then:

$L = \sqrt{2+\sqrt{L}} \Rightarrow L^2 = 2+ \sqrt{L} \Rightarrow L^4-4L^2-L+4=0 \Rightarrow (L-1)(L^3+L^2-3L-4)=0$

The limit can not be $1$ since the sequence is increasing and $a_1 > 1$, but I'm not able to find the roots of $L^3+L^2-3L-4$. ¿Is there another way to compute the limit? (It is a problem presented in a test so It would be convinient to solve it whitout using a calculator or computer)

Mittens
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MC2
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    Let a computer do it. Or put the coefficients in Cardano's formulas – NDB Aug 18 '23 at 13:16
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    @NDB It is a problem presented in a test so It would be convinient to solve it whitout using a calculator or computer – MC2 Aug 18 '23 at 13:22
  • @Mittens It is not quite the same problem. The problem I'm trying to solve was in a test and also I've found that It is on Rudin's Principals of Mathematical Analysis on chapter 3. But probably It has a mistake and instead of $\sqrt{2+\sqrt{a_n}}$ It should be $\sqrt{2+a_n}$ as in the problem that you suggested (since there is a soultion manual for Rudin's book by Roger Cooke and It considers $$\sqrt{2+{a_n} $$ instead of $\sqrt{2+\sqrt{a_n}}$) – MC2 Aug 18 '23 at 13:33
  • @GS2: I removed my comment. In any case, unless you enter Cardano or Tartaglia's formula, there are no simpler ways to get solutions other than numerical methods. In any event, the humble Descartes' sign rule shows that there is exactly one positive solution to the cubic equation. Furthermore, setting $a=\sqrt{L}$, one gets the equation $a^4-a-2=0$; Descartes rule also shows that there is exactly one positive solution. – Mittens Aug 18 '23 at 14:01
  • To be clear, this is$$\sqrt{2+\sqrt{\sqrt{2+\sqrt{\sqrt{\dotsb}}}}},$$full of double square roots. Is this what you want? – Akiva Weinberger Aug 18 '23 at 14:40
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    Observe that $a_n<2$, by the induction argument. Hence $a_{n+1}<\sqrt{2+\sqrt{2}}.$ So the limit is somewhere between $\sqrt{2+\sqrt[4]{2}}$ and the value above. – Ryszard Szwarc Aug 18 '23 at 14:44

3 Answers3

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The limit is:

The unique root of $x^3 + x^2 - 3x - 4=0$ that lies in the interval $[0,3]$.

Of course, in order for this solution to be valid, you must prove that this root exists and is unique. But that's easy using a little bit of calculus.

First, its value at $x=0$ is $-4$ and its value at $x=3$ is $+4$ and so, by the intermediate value theorem, there is at least one root in the interval $[0,3]$.

Next, its derivative is $3x^2+2x-3$. The derivative is $-3$ at $x=0$ and $30$ and $x=3$. Also, its second derivative is $6x+3$ which is positive on $[0,3]$. Therefore, the graph is concave up on $[0,3]$ and is decreasing starting at $x=0$, $y=-4$ and increasing at $x=3$, $y=4$ and so it crosses the $x$-axis only once in the interval $[0,3]$.

Lee Mosher
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Python solves it with Cardano's formulas

>>> from sympy import solve, symbols
>>> x = symbols('x')
>>> solve(x**3 + x**2 - 3 * x - 4, x)
[-1/3 + (-1/2 - sqrt(3)*I/2)*(sqrt(249)/18 + 79/54)**(1/3) + 10/(9*(-1/2 - sqrt(3)*I/2)*(sqrt(249)/18 + 79/54)**(1/3)), -1/3 + 10/(9*(-1/2 + sqrt(3)*I/2)*(sqrt(249)/18 + 79/54)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(249)/18 + 79/54)**(1/3), -1/3 + 10/(9*(sqrt(249)/18 + 79/54)**(1/3)) + (sqrt(249)/18 + 79/54)**(1/3)]

If the limit is the real root of the 3rd degree polynomial, there are no simpler ways to express the solution. Especially you won't avoid the $\left(\frac{\sqrt{249}}{18} + \frac{79}{54}\right)^{1/3}$ part, one way or another.

Gribouillis
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  • How do you now that there are no other positive solutions?Notice that the OP stated "It is a problem presented in a test so It would be convinient to solve it whitout using a calculator or computer". Other wise one can use out favorite numerical method and a a pocket calculator. – Mittens Aug 18 '23 at 14:03
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    @Mittens If there is a limit, it must satisfy the polynomial equation and be a real number. As this equation has only one real root, it must be this root. Now, you could prove by different means that the polynomial equation has only one real solution as in Lee Mosher's reply, but this does not give you the exact value. Here we use the computer only to apply the Cardano's formula for us. You could do it by hand. – Gribouillis Aug 18 '23 at 14:09
  • I understand that the limit satisfies a polynomial equation. The questions is, are there more solutions to the polynomial solution. The laser is no due to Descartes theorem, but you are not answering that question. You are just solving the equations with a computer package. Mathematica gives also solutions, in fact it gives the complex as well as the unique positive solution. But the OP stated clearly that use of calculators and computers was not allowed in the test. – Mittens Aug 18 '23 at 14:16
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    @mittens the criteria to see if a real cubic equation as 1 or 3 real roots is whether its discriminant is negative or positive. You can easily compute the discriminant by hand if you want https://en.wikipedia.org/wiki/Cubic_equation – Gribouillis Aug 18 '23 at 14:19
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Instead of proving (by calculus or algebraic methods) that the equation $$ x^3+x^2-3x-4\quad (*)$$ has a unique positive root, in order to determine the limit in OP, we will do the converse. The convergence of the sequence in OP will be applied to showing that the equation has a unique positive solution.

First observe that the equation has no roots $x$ less than or equal $\sqrt{3}$ as $$x(x^2-3)+(x^2-4)<0$$ Let $$f(x)=\sqrt{2+\sqrt{x}}$$ For positive $x$ the equation $(*)$ is equivalent to $x=f(x)$ as $x >\sqrt{2}.$ Since the sequence in $OP$ is convergent the equation has at least one positive solution. Assume by contradiction that $\sqrt{2}<r<s$ are solutions of $x=f(x).$ Then $$s-r=f(s)-f(r)={f(s)^2-f(r)^2\over f(s)+f(r)}\\ ={s-r\over (s+r) (\sqrt{s}+\sqrt{r} )}<s-r$$ which leads to a contradiction.

Ryszard Szwarc
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