I have a calculus final two days from now and we have a test example. There's a sequence question I can't seem to solve and hope someone here will be able to help. With $a_1$ not given, what are the possible values of it so that the sequence $a_{n+1}=\sqrt{3+a_n}$ will converge. If it does, what is the limit?

I have no clue what so ever on what doing here. I mean, I can't prove the sequence is monotone. I assume that $a_1$ $\ge $ -3 and can also approach infinity.

Any help is appreciated, Regards,

  • $\begingroup$ Let the sum go to $\infty$ and then treat the sum as a nested radical. $\endgroup$ – pidude Jul 2 '14 at 16:36

As @evinda rightly noticed, the limit must be $\frac12(1+\sqrt{13})$. For $a_1$ we can take an arbitrary number in $[-3,\infty)$. Note that $a_2$ will be nonnegative in any case and finally note that if $a_k\in [0,\frac12(1+\sqrt{13})]$, then $a_k\le a_{k+1}=\sqrt{a_k+3}\le \frac12(1+\sqrt{13})$ while if $a_k>\frac12(1+\sqrt{13})$, then $\frac12(1+\sqrt{13})<a_{k+1}=\sqrt{a_k+3}<a_k$; thus, the sequence is monotone starting from the second term and hence convergent.

  • $\begingroup$ So, $a_k$ can be an increasing or decreasing sequence depended on the value of $a_1$? (I miss clicked on Enter before) $\endgroup$ – Billy McGeen Jul 2 '14 at 17:04
  • $\begingroup$ What do you mean? $\endgroup$ – Vladimir Jul 2 '14 at 17:05
  • $\begingroup$ If I take $a_1\ge-3$, then $a_2=\sqrt{3+a_1}$ is nonnegative, right? $\endgroup$ – Vladimir Jul 2 '14 at 17:06
  • $\begingroup$ Indeed that is. $\endgroup$ – Billy McGeen Jul 2 '14 at 17:07
  • $\begingroup$ Next, if $0\le x \le\frac12(1+\sqrt{13})$, then $\sqrt{x+3}\ge x$, because $x+3>x^2$, right? $\endgroup$ – Vladimir Jul 2 '14 at 17:08

We want that the sequence $a_{n+1}$ converges, so $a_{n+1} \to l \in \mathbb{R} \Rightarrow a_n \to l$

Taking the limit $n \to +\infty$ at the relation $a_{n+1}=\sqrt{3+a_n}$ we get: $$l=\sqrt{3+l} \Rightarrow l^2=3+l \Rightarrow l=\frac{1}{2}(1-\sqrt{13}) \text{ OR } l=\frac{1}{2}(1+\sqrt{13})$$

As $\frac{1}{2}(1-\sqrt{13})$ is negative,we reject it,so the only possible limit of $a_n$ is $l=\frac{1}{2}(1+\sqrt{13})$.

  • 2
    $\begingroup$ what is the value of $a_1$ then? You did not fully answer his question $\endgroup$ – Varun Iyer Jul 2 '14 at 16:40
  • $\begingroup$ Yes, I was probably sure this is the limit, but what are the possible values of $a_1$? $\endgroup$ – Billy McGeen Jul 2 '14 at 16:44
  • $\begingroup$ I think it may just be $a_1 \geq -3$, you may be overcomplicating the problem $\endgroup$ – Trajan Jul 2 '14 at 16:47
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    $\begingroup$ Well, I'm almost sure that $a_1$ ≥ −3 but you have to prove it. I thought of either proving that the limit is not depended on a1 as long as it is bigger than -3, or either finding it. $\endgroup$ – Billy McGeen Jul 2 '14 at 16:54

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