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As said in the question, my task is to find the limit of the sequence as $n$ goes to $\infty$.

The sequence is defined as $a_1 = 1, a_{n+1} = \sqrt{2 + a_n}$ for all $n \in \mathbb{N}$.

Visually, the sequence looks like \begin{equation*} \{a_n\} = \left\{1, \sqrt{3}, \sqrt{2 + \sqrt{3}}, \sqrt{2 + \sqrt{2 + \sqrt{3}}}, ...\right\} \end{equation*} I am comfortable with solving limits, but this is the first time trying to solve a limit on an expression that is recursive (i.e., $\lim_{n \to \infty} \sqrt{2 + a_n}$, where $a_n$ continuously recurses). Hence, my brain doesn't really know where to go.

The closest thing I've found to solving this problem is a YouTube video that shows how to prove that the never ending $\sqrt{2 + \sqrt{2 + 2 + \sqrt{2 + 2 + \sqrt{2...}}}}$ converges to 2. However, in my problem, I am stuck with a $\sqrt{3}$ in the deepest root.

Any hints or pointers would be deeply appreciated.

Ang Ming Wen
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2 Answers2

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The sequence is defined as:

$$ a_{n+1} = \sqrt{2 + a_n}, \quad a_1 = 1. $$

Provided the limit exists, we may proceed. In this case, the monotonicity criterion for sequences states that a monotone, bounded sequence converges.

For a recursive sequence,

$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} =: a. $$

Hence,

$$ \lim_{n \to \infty} a_{n+1} = a = \sqrt{2 + a}. $$

Solving for $a$:

$$ a^2 = a + 2 $$

$$ \implies a^2 - a - 2 = 0. $$

Solving the quadratic equation:

$$ (a - 2)(a + 1) = 0. $$

Since $a > 0$, we conclude:

$$ a = 2. $$

Now, it remains to show, besides that the sequence is bounded and monotonous (do you want me to help you with that?):

$$ \forall \varepsilon > 0, \ \exists n_0 \in \mathbb{N}: \ \forall n \geq n_0 \implies |a_n - 2| < \varepsilon. $$

Rócherz
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Samuel
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    Thank you so much! Your explanation makes simple sense to me. Wasn’t able to think of it in this way. Thankfully, I was able to show the sequence being bounded and monotonous. I wanted to ask though - is this the requirement for the method you used? That is, setting $a = 2$ doesn’t represent until we can confirm the sequence converges - only then, 2 becomes the limit? Thanks so much again. – Ang Ming Wen Mar 03 '25 at 00:16
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    Glad I could help! Those are two distinct things you describe: The “fixed point“ approach (what I did to find the limit) calculates a limit. Proving convergence is necessary for that approach to hold. Also, while you prove that your sequence is bounded, you also encounter the fact that all elements of the sequence are positive. This strongly hints that a=-1 (which was another solution of the quadratic above) is no limit. If you try proving that, it won’t work out. Proving a = 2 is the limit, however, works out. – Samuel Mar 03 '25 at 12:44
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    TLDR: Yes, showing that the sequence converges is necessary before evaluating the limit via the fixed point approach. – Samuel Mar 03 '25 at 12:48
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    Appreciate the help and explanation so much @complexlogarithm. This will go a long way for me. – Ang Ming Wen Mar 04 '25 at 10:21
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If the sequence converge the limit is a fixed point of $x\mapsto\sqrt{2+x}=f(x)$ so $\ell=2$. Looking at the graph of this function you can see that with $a_0=1$ the sequence is increasing (the graph is above the bissectrice $y=x$).

You can prove it recurcively $a_{n+1}-a_n=\frac2{a_n+\sqrt{2+a_n}}>0$ or studying the sign of $f(x)-x$.

As $f$ is increasing $a_n<2\Longrightarrow f(a_n)=a_{n+1}<f(2)=2$. Increasing plus upper bounded$\Longrightarrow $ converge.

noradan
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