What is the following sequence classified as? I don't want to make anybody solve it, I just need to know where to begin looking to solve it. $$\alpha_1 = \sqrt{20}$$ $$\alpha_{n+1} = \sqrt{20 + \alpha_n}$$

I am suppose to prove that it converges to 5, however if I could just get a little terminology help it is more then appreciated!

Note: I updated the terminology, as well as give the initial value.


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    $\begingroup$ This is a "recursively defined" sequence, not a series. And for this to make sense, you need to give a starting value ($\alpha_1$, or $\alpha_n$). $\endgroup$ – Arturo Magidin Apr 29 '11 at 15:33
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    $\begingroup$ The word "moron" is excessive. People are not always familiar with the correct mathematical terminology. The important thing is that you fixed the mistake. $\endgroup$ – Asaf Karagila Apr 29 '11 at 15:59
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    $\begingroup$ We all make mistakes; it is good that you now have rectified your post, and so there is no need to beat yourself up for your previous mistake. $\endgroup$ – J. M. is a poor mathematician Apr 29 '11 at 16:05
  • $\begingroup$ @Asaf: It sure is! :-) $\endgroup$ – Aryabhata Apr 29 '11 at 16:12
  • $\begingroup$ Moron: Don't be an ass. ;-) $\endgroup$ – Asaf Karagila Apr 29 '11 at 16:15

First, it's not a series, it's a sequence. Fixed in the original.

Second, it's a recursively defined sequence.

A sequence is "recursively defined" if you specify some specific values and then you explain how to get the "next value" from the previous one; much like induction. Here, you are saying how to get the "next term", $\alpha_{n+1}$, if you already know the value of the $n$th term, $\alpha_n$.

Once you know the first value, then the sequence is completely determined by that first value and the "recurrence rule" $\alpha_{n+1}=\sqrt{20+\alpha_n}$.

Now some hints:

  • Show the sequence is increasing.
  • Show the sequence is bounded.
  • Conclude the sequence converges.
  • Once you know it converges, take limits on both sides of the recursion to try to figure out what it converges to.
  • $\begingroup$ I apologize about that, I updated the original post. $\endgroup$ – Addo Solutions Apr 29 '11 at 15:51

In addition to the excellent hinting of Arturo, I say that it might be useful to consider the intuitively-inappropriate statement that $ x = \sqrt{20 + x} $, or rather that $x^2 = 20 + x$.

To be clear, the existence of a solution to this statement does not imply the existence of a solution to your recurrence, but after following Arturo's hints...

Good luck!


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