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Evaluate

$$\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$$

I need some help with this question because I have no idea what is going on and help would be greatly appreciate :)

Joe
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2 Answers2

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$\sqrt{6+\sqrt{6+\sqrt{6+\dots}}} = \lim x_n$ where $x_{n+1} = \sqrt{6+x_n}$ and $x_0 \ge 0$.

As others have explained, if this sequence converges, then it converges to $3$.

To prove that it does converge, prove this:

  • If $x_0 < 3 $ then the sequence is increasing and bounded above by $3$.

  • If $x_0 > 3 $ then the sequence is decreasing and bounded below by $3$.

In both cases, you get a monotone bounded sequence, and so it converges.

lhf
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  • Isn't it worth mentioning that $x_0=\sqrt{6}$? (+1), by the way. – Vincenzo Oliva Jun 23 '15 at 12:09
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    @VincenzoOliva, this works regardless of $x_0$. It is natural to take $x_0=0$ but who knows what lies at the end of those ellipsis in $\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$ ? :-) – lhf Jun 23 '15 at 12:16
  • I know. :-) I meant in this case, to mention that the sequence in the OP is in fact increasing. Indeed $0$ and $\sqrt{6}$ lead to the same thing.

    But yeah, I guess that's unimportant, my bad :D

     – Vincenzo Oliva Jun 23 '15 at 12:25
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Put $$ x=\sqrt{6+\sqrt{6+\sqrt{6+\dots}}} $$ squaring you get $$ x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}=6+x\;\;\;. $$ Thus you simply have to solve $$ x^2-x-6=0\;\; $$ which has two solutions: $x=-2$ and $x=3$. But $x$ is clearly positive, thus you can conclude that $$ \sqrt{6+\sqrt{6+\sqrt{6+\dots}}}=3\;. $$

Joe
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