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The main question goes like this:

$$x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$$

Comment on $x$.

There are options given as well:
A) $x$ is an irrational number
B) $2<x<3$
C) $x=3$
D) None of these

My approach: So I try to square the term first. Pretty obvious, it's not going to get me anywhere, I think. I think $x$ is irrational, but I don't have any solid reason. Please help me out.

Michael Rozenberg
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  • This can be generalized into a formula. For reference, see this: http://math.stackexchange.com/questions/515253/how-to-prove-sqrt5-sqrt5-sqrt5-sqrt5-sqrt5-sqrt5-sqrt5-cdots/ – Frank Jul 03 '16 at 04:38

4 Answers4

11

$$x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}$$$$x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=6+x$$$$x^2-x-6=(x+2)(x-3)=0$$$$x=-2~or~3$$ $x=-2$ is obviously a bad answer

Therefore, $x=3$

To sate the appetite for rigor of the commenters below: Yes, the series does converge. This is because $3<\sqrt{6+y}<y$ if $y>3$ and $y<\sqrt{6+y}<3$ if $-6<y<3$.

Nathaniel B
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    Nice work, thanks a lot pal! –  Jul 03 '16 at 04:29
  • This solution is total wrong! Why this series is a convergent series? – Michael Rozenberg Jul 03 '16 at 04:46
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    @MichaelRozenberg $3<\sqrt{6+x}<x$ if $x>3$ and $x<\sqrt{6+x}<3$ if $-6<x<3$ – Nathaniel B Jul 03 '16 at 04:53
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    @MichaelRozenberg The solution isn't "wrong." If anything it relies on the hypothesis that the sequence, $a_{n+1} = \sqrt{6 + a_n}$, converges. This hypothesis is true, but the poster didn't prove this. You could claim it's incomplete, but it definitely isn't wrong. Also, I would add that based on the OP's comments on your answer, he hasn't taken a single variable calculus course meaning that your answer isn't very enlightening. For a pre-calculus student some hand-waving in the answer is okay. – bthmas Jul 03 '16 at 04:59
  • Dear @bthmas ! You can not assume a convergence without proof and it does not depend on age. By the way, the post was fixed. It seems not good. – Michael Rozenberg Jul 03 '16 at 07:44
  • @MichaelRozenberg The poster wasn't assuming convergence without proof. It is a trivial task to show that this sequence converges and something that can be omitted when the OP wouldn't be able to follow said proof anyways. First-year calculus students freely use the fundamental theorem of calculus when a truly rigorous proof is far out of reach for them to grasp. – bthmas Jul 03 '16 at 07:57
6

Let $a_{n+1}=\sqrt{6+a_n}$, $a_1=\sqrt{6}$.

We'll prove that $a_n<3$.

Since $a_1<3$, it remains to prove that $a_{n+1}<3$ for $a_n<3$.

Indeed, $a_{n+1}=\sqrt{a_n+6}<\sqrt{3+6}=3$.

Id est, $a_n<3$.

By another hand, $a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{(3-a_n)(2+a_n)}{\sqrt{a_n+6}+a_n}>0$,

which says that there is $\lim\limits_{n\rightarrow+\infty}a_n$.

Let $\lim\limits_{n\rightarrow+\infty}a_n=A$.

Hence, $\lim\limits_{n\rightarrow+\infty}a_{n+1}^2=\lim\limits_{n\rightarrow+\infty}a_n+6$ or $A^2=A+6$ or $A=3$, which gives the answer.

Michael Rozenberg
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    What's lim? I'm still 15... –  Jul 03 '16 at 04:42
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    @user351709 See here https://en.wikipedia.org/wiki/Convergent_series – Michael Rozenberg Jul 03 '16 at 04:50
  • Possibly needs a bit more explanation about how the sequence you define relates to the "expression" $\sqrt{6+\sqrt{6+\dots }}$ – Conrad Turner Jul 03 '16 at 06:54
  • @MichaelRozenberg Your first line's $;a_{n+1};$ is, I think, missing a square root on the right side. – DonAntonio Jul 03 '16 at 07:33
  • @ConradTurner I can't see how this definition needs anything else: the given sequence's clearly defined as a recursive one. What else is needed? – DonAntonio Jul 03 '16 at 07:35
  • @Joanpemo A statement that the string $\sqrt{6+\sqrt{6+\dots}}$ means nothing but the limit of this sequence (and maybe an explanation about why?). – Conrad Turner Jul 03 '16 at 10:08
  • @ConradTurnerThank you, but I don't get your point. The answerer is defining the expression $;\sqrt{6+\sqrt{6...}};$ in a recursive manner and is thus turning the expression into a sequence. Perhaps I'm missing something but I can't see what isn't clear here. I also don't understand how can you write "a statement that the string... means nothing but the limit of this sequence..." , so the string means nothing or is it a sequence? – DonAntonio Jul 03 '16 at 10:15
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Solution C is correct.

Since $x=\sqrt{6+\sqrt{6+\cdots }}$ we have that:

$$ x=\sqrt{6+x}$$

This reduces to the quadratic equation $x^2-x-6=0$ which has solutions $x=-2$ or $x=3$. $x=-2$ is not a solution since $-2\neq \sqrt{6-2}=\sqrt{4}=2$. Thus $x=3$.

M47145
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    Oh, great idea, thanks a bunch! –  Jul 03 '16 at 04:29
  • This solution is total wrong! Why this series is a convergent series? – Michael Rozenberg Jul 03 '16 at 04:47
  • @MichaelRozenberg Twice in this thread you call a solution "totally wrong", and I think that is...yes, wrong. First, the solution is right, though one could say one or two things about the justification. Second, answers could be seeking to give only part of hte soltuion, not the hole thing. Third, the question's solution methods here is about a sequence defined recursively, not a series . – DonAntonio Jul 03 '16 at 06:27
  • @Joanpemo I think what MichaelRozenberg is trying to get at is that these posts are not establishing what the posters are taking $\sqrt{6+\sqrt{6+\dots }}$ to mean. As these stand they are equivalent to "proving" $0.99..=1$ by jiggery-pokery with applying arithmetical operations to non-terminating decimals without reference to what $0.99..$ means (as a limit of a series) – Conrad Turner Jul 03 '16 at 06:48
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    @ConradTurner Thank you. That's a lot to deduce from one line which doesn't say anything at all about "taking this as this or that", but talks of convergence matters of "a series" . What I see is that the answeres are referring implicitly to the sequence $$x_1=\sqrt 6,,,,x_2=\sqrt{6+\sqrt 6},,...,x_n=\sqrt{6+x_{n-1}}$$ and then assuming this sequence's limit exists, they just use arithmetic of limits. It is very different, in my opinion, of the proof $;0.\overline9=1;$ using an infinite convergent geometric series. – DonAntonio Jul 03 '16 at 07:31
  • @Joanpemo The kind of proof that $0.99\dots=1$ I had in mind were of the type: $10\times 0.99\dots =9.99\dots$ then $10\times 0.99\dots -9=0.99\dots$, so after rearranging $9 \times 0.99\dots=9$ so $0.99\dots=1$. That is without considering what $0.99\dots$ or in this case $\sqrt{6+\sqrt{6+\dots}}$ is/are supposed to mean. – Conrad Turner Jul 03 '16 at 10:16
0

An empirical approach:

Starting fom $0$, the first iterates (truncated) are

$$2.449,2.907,2.984,2.997,3.000$$

Hence you can conjecture the value $3$, and easily prove that $3$ is indeed a possible solution as $$\sqrt{6+3}=3.$$

Unfortunately, this is technically inconclusive as it doesn't rule out the existence of another solution or the possibility of non-convergence.

Now, studying convergence, we can write

$$3-\lambda\epsilon<\sqrt{6+(3+\epsilon)}<3+\lambda\epsilon$$

which simplifies to

$$-6\lambda+\lambda^2\epsilon<1<6\lambda+\lambda^2\epsilon.$$

This shows that sufficiently close to the value $3$, we have linear convergence with a common ratio approaching $1/6$. This is confirmed by $$\sqrt{6+3.000006}\approx 3.000001$$ with an excellent approximation.