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I was studying about solving expressions of the form $\sqrt{6+\sqrt{6+\sqrt {6+\cdots}}}$

I know how to solve this, but how can this equation have a constant variable if it is continued indefinitely?

I know upto limits and basic derivatives. Thanks

Martin Sleziak
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Soham
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    I think you're going to need to clarify this question quite a bit before you get the answer you want. – Jerry Guern Oct 06 '15 at 17:21
  • Do you mean an infinite number of iterations or do you mean an infinity in the expression? I pretty sure its the former – Chinny84 Oct 06 '15 at 17:22
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    Whatever reason you have put the $\infty$ in there for, it shouldn't be there. – Chappers Oct 06 '15 at 17:22
  • @Chappers-These days books are teaching wrong things.It's given in my book. – Soham Oct 06 '15 at 17:29
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    Didn't you rather mean $ \cdots +\sqrt{6 + \ldots + \sqrt{6+\sqrt{6+\sqrt{6}}}}$ where the evaluation begins in the innermost square-root? And then the partial evaluations should oscillate between that value when you have evaluated up to some $6$ and when you have evaluated up to one $\sqrt{;}$ ... – Gottfried Helms Oct 06 '15 at 17:32
  • What do you mean by "a constant variable"? – JRN Oct 07 '15 at 04:30

2 Answers2

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Define a sequence $a_1 = \sqrt 6, a_2 = \sqrt{6 + \sqrt{6}} = \sqrt{6 + a_1}$ and recursively $a_{n+1} = \sqrt{6 + a_n}$ for all $n \geq 1$.

Now show that the sequence $(a_n)$ is bounded above and increasing. Then the sequence has a limit.

Call that limit $L$. As $L = \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \sqrt{6 + a_n}$ we have

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

From this you can find the value of $L$.

We can also generalize this to make sense of expressions such as

$$\sqrt{a + \sqrt{a + \sqrt{a + \cdots}}}$$

Simon S
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  • Of course, the initial value $a_1$ need not be $\sqrt{6}$ to arrive at this result. I posted an answer that examines the various cases and would like to have your thoughts if you have the time. ;-)) – Mark Viola Oct 06 '15 at 17:47
  • By the way ... +1 – Mark Viola Oct 06 '15 at 18:11
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If we define the sequence $x_n$ through the recurrence relationship $x_{n+1}=\sqrt{6+x_n}$. There are three cases for values of $x_1$.

CASE $1$: $x_1=3$

Obviously, if $x_1=3$, then $x_n=3$ for all $n>1$.

CASE $2$: $-6\le x_1<3$

If $-6\le x_1<3$, then it is easy to see inductively that $0<x_{n}<3$ for all $n > 1$. Additionally, it is easy to see that $x_{n+1}-x_n=\sqrt{6+x_n}-x_n=\frac{(3-x_n)(2+x_n)}{\sqrt{6+x_n}+x_n}>0$ and thus $x_n$ increases monotonically. Therefore we have

$$\lim_{n\to \infty}x_{n+1}=\lim_{n\to \infty}x_n=\sqrt{6+\lim_{n\to \infty}x_n}\implies \lim_{n\to \infty}x_n=3$$

CASE $3$: $3<x_1$

If $3<x_1$, then it is easy to see inductively that $0<x_{n}>3$ for all $n\ge 1$. Additionally, it is easy to see that $x_{n+1}-x_n=\sqrt{6+x_n}-x_n=\frac{(3-x_n)(2+x_n)}{\sqrt{6+x_n}+x_n}<0$ and thus $x_n$ decreases monotonically.

Therefore we have $$\lim_{n\to \infty}x_{n+1}=\lim_{n\to \infty}x_n=\sqrt{6+\lim_{n\to \infty}x_n}\implies \lim_{n\to \infty}x_n=3$$

Mark Viola
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  • In response to your request: This all looks fine. The expression as expressed in the OP is ill defined and so it's not a bad idea to think about a few different alternatives of a sound definition and show they all give the same answer. I chose what I thought of as the most natural definition. – Simon S Oct 06 '15 at 17:56
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    @SimonS Yes, the problem is ill-posed. This is an issue that is often misunderstood. There is no rigorous meaning to $\sqrt{6+\sqrt{6+\cdots}}$. It is a matter of inside-out versus outside-in. – Mark Viola Oct 06 '15 at 18:04