Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,

share|cite|improve this question
Let the sum go to $\infty$ and then treat the sum as a nested radical. – pidude Jul 2 '14 at 16:36
up vote 9 down vote accepted

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.

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

share|cite|improve this answer
what is the value of $a_1$ then? You did not fully answer his question – Varun Iyer Jul 2 '14 at 16:40
Yes, I was probably sure this is the limit, but what are the possible values of $a_1$? – Billy McGeen Jul 2 '14 at 16:44
I think it may just be $a_1 \geq -3$, you may be overcomplicating the problem – Permian Jul 2 '14 at 16:47
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. – Billy McGeen Jul 2 '14 at 16:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.