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The recursive sequence given by

$ s_1 = \sqrt{2} $

$ s_{n+1} = \sqrt{2 + s_n} $

seems to converge to $ 2 $

It's easy to prove that this sequence indeed converges and that $ 2 $ is a upper bound for that sequence, but how to prove that it really does converge to 2?

thanks in advance.

Vinicius L. Deloi
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    Brute force way: over $\mathbb{R}^+$, $f(x)=\sqrt{2+x}$ is a contraction since $|f'(x)|<1$. By the Banach fixed point theorem the sequence ${s_n}_{n\geq 1}$ converges to the only solution of $f(x)=x$ in $\mathbb{R}^+$. – Jack D'Aurizio Mar 20 '18 at 03:09
  • Another idea is to show that the sequence is Cauchy by subtracting $s_{n+1}-s_n$, and showing this difference is arbitrarily small when $n$ gets large. – BSplitter Mar 20 '18 at 03:20
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    The problem is not to prove that the sequence converges but to prove it converges to the limit two, but thanks anyway. – Vinicius L. Deloi Mar 20 '18 at 03:21
  • A fun way to get this result is to let $r_{n+1} = \sqrt{4-2^{1-n}}$ and show inductively that $r_{n+1} \leq s_{n+1}$ for all $n\geq 1$. Then use squeeze theorem and the fact that $r_n \rightarrow 2$ to get the desired result. – Alex Jones Mar 20 '18 at 03:54

3 Answers3

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You know that the sequence has a limit $L$

Let $ n\to \infty$ and apply the result to $$s_{n+1} = \sqrt{2 + s_n}$$

You get $$ L= \sqrt{2 + L}$$

Upon solving for $L$, we get two solutions. $$ L= 2$$ and $$ L=-1$$ $L=-1$ is out of question because all our terms are positive.

Thus $$L=2$$

Mohammad Riazi-Kermani
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Suppose that you do not know that the square root is continuous (or even the sequential characterization of continuity). Then we can do the following to show that if it converges to $2$:

We can rewrite the inductive step as $s_{n+1}^2 = 2 + s_n$. Since these two sequences are the same, their limits will be the same. Suppose $s_n \to L$. Then taking the limit of both sides of the equation, we get that $L^2 = 2 + L$. Solving, we get that either $L=-1$ or $L = 2$. Since the sequence is always positive, $L = 2$.

Jonathan Dunay
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Let $r_{n+1} = \sqrt{4-2^{1-n}}$. Then, $r_1 = \sqrt{2} \leq \sqrt{2} = s_1$.

Further, assume that $r_{n+1} = \sqrt{4-2^{1-n}} \leq \sqrt{2+s_n} = s_{n+1}$.

Then, $s_{n+2} = \sqrt{2+s_{n+1}} = \sqrt{2+\sqrt{2+s_n}} \geq \sqrt{2+\sqrt{4-2^{1-n}}}$ by hypothesis.

Since $n \geq 0 \geq -1$,

$1-n \geq -2n \Rightarrow 2^{1-n}\geq 2^{-2n} \Rightarrow 4+2^{2-n}-2^{1-n} \geq 4+(2^{-n})^2 \Rightarrow 4-2^{1-n}\geq (2-2^{-n})^2$.

Thus, $\sqrt{2+\sqrt{4-2^{1-n}}} \geq \sqrt{2+2-2^{-n}} = \sqrt{4-2^{-n}} = r_{n+2}$.

Thus, for all $n \geq 0$, $r_{n+1} \leq s_{n+1}$.

By squeeze theorem, since $r_{n+1} \rightarrow 2$ and $s_{n+1} \rightarrow L \leq 2$, $s_{n+1}\rightarrow 2$.

Alex Jones
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