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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.

Thanks!

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    I am glad the math people have a good sense of humor :) I just know it is really annoying when n00b's make stupid mistakes and you have to write a really long message to explain why (I am not great with math, but give me software to write, and I am told I am less n00bish :). Thanks for all the help guys!2011-04-30

2 Answers 2

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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.
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    I apologize about that, I updated the original post.2011-04-29
2

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!