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Let $a_n$ be the sequence in $\mathbb{R}$ given by:

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

Show that this sequence converges and find the limit.

I have already proven that this sequence is increasing and bounded by $2$, so the limit exists. Now, when I'm trying to find the limit, say $L$, I get that $L$ satisfies the equation: $$L^2 -2 - \sqrt{L}=0, $$ but I don't actually know how to solve this equation, so I would like to know if there is an easier way to find the limit, and in case I must solve the last equation, how can I solve it?

Thanks in advance!

Taroccoesbrocco
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user392559
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  • I am not aware of any other way of obtaining the limit. Solving this might be bit of an overkill, but here is something that might help. – onetimething Mar 22 '18 at 05:07
  • I wouldnt worry too much about it. Sometimes a teacher will ask you to prove convergence but not expect you to find the answer. Its not always easy to find an exact answer, sometimes its important to just be able to show that there is or isnt one to find. – CogitoErgoCogitoSum Mar 22 '18 at 05:20
  • If you have to crank out a numeric value, you can just use a reiterative process. Just type $\sqrt{2}$ into your calculator as your $a_1$, then repeatedly enter $\sqrt{2+\sqrt{ANS}}$. Do that a few times and you have a decent answer. Otherwise, to find an actual algebraic expression for this particular problem will require solving a cubic, which is above and beyond the scope of the course, I think. – CogitoErgoCogitoSum Mar 22 '18 at 05:22
  • Also, https://math.stackexchange.com/questions/2585780/show-that-the-sequence-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2/ – rtybase Mar 22 '18 at 09:37
  • Also, https://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq – rtybase Mar 22 '18 at 09:38
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    @rtybase None of these are duplicates. This question has an added square root. – Arnaud D. Mar 22 '18 at 13:24
  • @ArnaudD. but techniques are effectively the same ... – rtybase Mar 22 '18 at 15:47
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    @rtybase: Techniques are commom to thousands of solutions. But that is not a reason to claim that a question is a duplicate of a previous—distinctly different—question, nor is it a reason to close the question. Moreover, the solution to the present question involves solving a quartic, while the earlier question only required the solution of a quadratic. – John Bentin Mar 31 '18 at 10:35
  • @JohnBentin ok, I will start flooding MSE with questions asking for solutions for different polynomials in $\mathbb{Z}[x]$ (to make is simpler) of degree $\geq 5$, since, according to you, these will be different questions. Deal? – rtybase Mar 31 '18 at 11:03
  • To clarify my point, there is a common technique of solving convergence of ${a_n}$ from the function generating the sequence $a_{n+1}=f(a_n)$, by studying the function $f(x)$. Once the existence of the limits is identified, it is resolved by solving $L=f(L)$. If $f(x)$ is an equation (e.g. polynomials) and OP can's solve it, the question should be tagged adequately (e.g. "polynomials"). As with the sequence, I still think it's a dup. – rtybase Mar 31 '18 at 11:21

3 Answers3

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Let $M = \sqrt{L}$ then $$M^4-M-2=0,$$ with $M=-1$ being the obvious solution. You need to divide the LHS by $M+1$ and you get a cubic which is solvable exactly with Cardano's Formula.

If you are lazy, just plug into Wolfram Alpha to get $$ M = \frac13 \left(1 - 2 \left(\frac{2}{47 + 3 \sqrt{249}}\right)^{1/3} + \left(\frac{47 + 3 \sqrt{249}}{2}\right)^{1/3}\right) $$

gt6989b
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Look for a fixed point such that $x = a_{n+1} = a_n$:

\begin{align*} x &= \sqrt{2 + \sqrt{x}} \\ x^4 - 4x^2 - x + 4 &= 0 \\ (x-1)(x^3 + x^2 - 3x - 4) &= 0 \\ \end{align*}

The $x=1$ root is out of range of the sequence, so the only remaining fixed point is the root to $x^3 + x^2 - 3x - 4 = 0$ which is approximately $x \approx 1.8311772$.

clay
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You are on the right track. The square root in your equation for $L$ is awkward. So let's get rid of it by writing $L=x^2$; we can recover $L$ from $x$ at the end by squaring. The equation to solve becomes$$f(x):=x^4-x-2=0.$$The positive solution can be written down, after much labour, as a complicated expression involving nested radicals. However, if you want to know what it is numerically, the easy way is to use Newton–Raphson iteration. This method has the lovely property of quadratic convergence, and only a few iterations are needed to reach the limit of precision of a hand calculator. The basic formula is $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\quad(n=0,1,...),$$with a given starting value of $x_0$. In the present case, $f'(x)=4x^3-1$, and the formula simplifies to$$x_{n+1}=\frac{3x_n^4+2}{4x_n^3-1}.$$Let's start with $x_0=\frac43$ as a reasonably simple first estimate of the root (which is actually an underestimate by about $1.5$%). The formula gives $x_1=310/229$, which is out by less than $4$ parts in $10,000$. The error in $x_2$ is below $4$ parts in $10,000,000$, and the error in $x_3$ is smaller than an ordinary calculator will show.

John Bentin
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