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QUESTION:
Show that the sequence {$\sqrt{5},\sqrt{5+\sqrt{5}},\sqrt{5+\sqrt{5+\sqrt{5}}},\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5}}}},....$} is convergent and it converges to $\left(\frac{1+\sqrt{21}}{2}\right)$.

MY ATTEMPT:
The sequence takes the form of the recurrence $x_n=\sqrt{x_{n-1}+5}$. But neither can I show it to be monotonic increasing nor bounded. Once I have shown it to be convergent, I know how to find and show the limit. I have successfully done it too. But I cannot prove the convergence.

ONLY HINTS required.

P.S. Do not use Cauchy principle or any complicated test. I want the answer to be based on monotonicity and boundedness. For the problem belongs to that chapter only.

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  • $\begingroup$ $\sqrt{a}<\sqrt{b}$ if and only if $a< b$ $\endgroup$
    – Chinny84
    Commented Nov 10, 2015 at 7:34
  • $\begingroup$ @Chinny84 How do I use this to prove monotonicity for large values of n? Induction perhaps? $\endgroup$ Commented Nov 10, 2015 at 7:36
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    $\begingroup$ Hint: If $x_0<x_1$, can you compare $x_1$ and $x_2$? Which one is greater? $\endgroup$
    – Did
    Commented Nov 10, 2015 at 7:38
  • $\begingroup$ @Did Your hint points to induction, I believe? $\endgroup$ Commented Nov 10, 2015 at 7:44
  • $\begingroup$ INDUCTION and again INDUCTION $\endgroup$
    – Bumblebee
    Commented Nov 10, 2015 at 8:46

2 Answers 2

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Hints:

  • use induction
  • if $a_n \gt a_{n-1}$ what can you say about $\sqrt{5+a_n}$ compared with $\sqrt{5+a_{n-1}}$?
  • if $a_n \lt 4$ for example, what can you say about $\sqrt{5+a_n}$?
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  • $\begingroup$ First bullet is for monotonicity, right? Now for 2nd bullet, when I know that $a_n>a_{n-1}$ why do I need to prove the root part? And for 3rd bullet, how do I say that $a_n<4$....intuition? $\endgroup$ Commented Nov 10, 2015 at 7:54
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    $\begingroup$ You might want to show $a_{n+1} =\sqrt{5+a_n} \gt a_n$ as part of a proof by induction of monotonicity. Similarly you might want to show $a_{n+1} =\sqrt{5+a_n} \lt 4$ as part of a proof by induction of boundedness. As initial steps, you will need to show $\sqrt{5+\sqrt{5}} \gt \sqrt{5}$ and similarly $\sqrt{5} \lt 4$. $\endgroup$
    – Henry
    Commented Nov 10, 2015 at 7:58
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Have you thought about using functional approach? consider $f(x) = \sqrt{x+5}$, and show $f$ is increasing. This takes care of the monotonicity of $a_n$'s, and you can show $a_n < 3$ by induction.

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