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For positive integer $n$, $$x_n=\sqrt {6+\sqrt{6+\sqrt{6+...+\sqrt 6}}}$$ where $6$ is written $n$ times.

How can we find the $x _\infty$ ?

I coded a program any found that $x _\infty$ would be equal to $3$. So

$$x _\infty =3$$

But I need to know how can we prove it. Or simply how can we find $x _\infty$?

Tharindu Sathischandra
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    The sequence of partials is increasing. An easy induction shows it is bounded above, for example by $3$. So the limit exists. Now it is just a computation, $x=\sqrt{6+x}$ – André Nicolas Mar 24 '16 at 04:28

3 Answers3

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HINT:

If $y=\sqrt{n+\sqrt{n+\cdots}},y=\sqrt{n+y}$

$\implies y^2=n+y$ and $y\ge0$

For convergence see Convergence of nested radicals and On Infinite Radicals

Community
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lab bhattacharjee
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  • See https://www.math.washington.edu/~morrow/336_11/papers/brian.pdf – lab bhattacharjee Mar 24 '16 at 04:33
  • Rigorously, one needs a seed value and a well-defined recurrence relationship. Here, we can say the seed is $y_0$ with $y_k=\sqrt{n+y_{k-1}}$. We arrive at the same conclusion, but I thought that this comment might be useful inasmuch as it codifies clearly the recurrence. – Mark Viola Mar 24 '16 at 05:16
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By definition, it is clear all $x_n > 0$. Notice for any $n > 1$, $$|3 - x_{n}| = |3 - \sqrt{6+x_{n-1}}| = \left|\frac{3-x_{n-1}}{3+\sqrt{6+x_{n-1}}}\right| < \frac13 |3-x_{n-1}|$$ This leads to for all $n > 0$, $$|3-x_n| \le \frac{1}{3^{n-1}}|3 - x_1| = \frac{1}{3^{n-1}}|3-\sqrt{6}|$$ Since the RHS tends to $0$ as $n \to \infty$, $\lim_{n\to\infty} x_n$ exists and equals to $3$.

achille hui
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Take $n \to \infty$, then we can write your expression as $$y = \sqrt{6 + \sqrt{6 + \sqrt{6 \dots}}}$$ $$\implies y = \sqrt{6 + y}$$ $$\implies y^2 = 6 + y$$ $$\implies y^2 - y - 6 = 0$$ $$\implies y = \frac{1 \pm \sqrt{1 + 24}}{2}$$ $$\implies y = \frac{1 \pm 5}{2} $$ $$\implies y = 3, -2$$ But we know that $y \ge 0$ since the RHS in of the original expression is $ > 0$ $$\implies y = 3$$ $$\text{Q.E.D.}$$

Banach Tarski
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