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If $s_1=\sqrt 2$, and $$s_{n+1}=\sqrt {2+\sqrt{s_n}}$$

Prove that ${s_n}$ is converging, and that $s_n<2$ for all $n=1,2\ldots$

My attempt to use mathematical induction: if $s_n<2$ then I use $s_{n+1}=\sqrt {2+\sqrt s_n}$, to prove $s_{n+1}<\sqrt {2+2}=2$. I guess $s_n$ is an increasing sequence, I can use the $s_n$ is bounded and increasing, and then prove that $s_n$ converges.

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marked as duplicate by rtybase, stressed out, user21820, Trevor Gunn, StubbornAtom Feb 2 at 22:37

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    $\begingroup$ Jackson.All is in place.Just work it out clearly.+ $\endgroup$ – Peter Szilas Feb 2 at 10:05
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    $\begingroup$ Thanks a lot for you answer my question ,I figure it out $\endgroup$ – jackson Feb 2 at 10:06
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    $\begingroup$ also covered here $\endgroup$ – rtybase Feb 2 at 10:28
  • $\begingroup$ @rtybase how you search for that I have searched before I asked but I still don’t find the answer.... $\endgroup$ – jackson Feb 2 at 10:28
  • $\begingroup$ I have dealt with this problem before ... just from the top of my head. $\endgroup$ – rtybase Feb 2 at 10:29
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To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.

Lemma: $s_n$ is bounded

Base case: $n=1$

$$s_1= \sqrt{2} <2$$

Suppose now for arbitrary $k$

$$ s_k< 2$$

Now observe as the inductive step: $$ s_{k+1}=\sqrt{2 + \sqrt{s_k}}< \sqrt{2+2}=2$$

By the principle of mathematical induction we have found that $2$ is a bound for all $n\in \mathbb N$.

Lemma: $s_n$ is increasing

Base case: $$s_2-s_1>0$$ $$\sqrt{2 + \sqrt{\sqrt{2}}}- \sqrt{2}>\sqrt{2 + 0}- \sqrt{2}=0 $$

Now suppose for arbitrary $k$ we have: $$ s_{k}-s_{k-1}>0 \implies s_k > s_{k-1}\implies \sqrt{s_k}> \sqrt{s_{k-1}}$$

We now make our inductive step: $$ s_{k+1}-s_k=\sqrt{2 + \sqrt{s_k}}-\sqrt{2 + \sqrt{s_{k-1}}}>\sqrt{2 + \sqrt{s_{k-1}}}-\sqrt{2 + \sqrt{s_{k-1}}}=0$$ Hence we have an increasing sequence for all $n \in \mathbb N$ this completes the proof and our sequence is indeed convergent $\square$.

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Your idea is perfect. The monotone convergence theorem guarantees convergence of a sequence if you can show that the sequence is increasing and bounded above (or decreasing and bounded below). To get that your sequence is increasing, the trick is to use strong induction:

The base case is easy, so I'll leave that to you. Next, suppose that each subsequent term is larger than the previous for each of the first $k$ terms. We want to show $s_{k+1} > s_k$, and because we used strong induction, we can assume $s_{k} > s_{k-1}$, which gives the inequality $s_{k+1} = \sqrt{2 + \sqrt{s_k}} > \sqrt{2+\sqrt{s_{k-1}}}$. And what is $\sqrt{2 + \sqrt{s_{k-1}}} \ $?

This argument is implicitly using the fact that $\sqrt{\cdot}$ is an increasing function, but this is fairly easy to prove.


If you wanted to actually find the limit of the sequence, first note that $\{s_{n+1} \}$ is a subsequence of $\{s_n\}$. If $ \{s_n \}$ converges, any subsequence also converges to the same value. Taking a limit of both sides of that equality, we get:

$$\lim(s_{n+1}) = \lim \left( \sqrt{2 + \sqrt{s_n}} \right)$$

Applying limit laws, and recognizing that $\sqrt{\cdot}$ is a continuous function$^\dagger$, we'll have:

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

We know $\{s_n\}$ converges to some $L \in \mathbb{R}^+$ and $\{s_{n+1} \}$ to the same, so the above reduces to:

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

Solving for $L$, we find that it is a root of the quartic $L^4 - 4L^2 - L + 4$. Notice that $1$ is a root, so we can factor out $(L-1)$, reducing the problem to finding roots of the cubic $L^3 + L^2 - 3L - 4$. You can find its three roots using the cubic formula (two of them will have nonzero imaginary part, so our $L$ will be the single real root). The cubic formula is a mess, so you can also arrive at an arbitrarily good approximation for $L$ using Newton's method. Either way, we get $L \approx 1.8312$.


$^\dagger$We're applying the fact that, if $f$ is a continuous function and $\{a_n\}$ a convergent sequence in the domain of $f$, then $a_n \to M \implies f(a_n) \to f(M)$.

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  • $\begingroup$ in fact I want to know how to prove $s_n$ is increasing😂😅 $\endgroup$ – jackson Feb 2 at 9:48
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    $\begingroup$ I know the fact you add,but I still don’t understand why $s_n$ is increasing $\endgroup$ – jackson Feb 2 at 9:53
  • $\begingroup$ Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas? $\endgroup$ – Kaj Hansen Feb 2 at 9:57
  • $\begingroup$ There we go, lol $\endgroup$ – Kaj Hansen Feb 2 at 9:58
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    $\begingroup$ Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck. $\endgroup$ – Kaj Hansen Feb 2 at 10:04
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It is a fixed point iteration of the form $s_{n+1}=g(s_n)$. You can easily check that $g([0,2]) \subset [0,2]$ an so, if you start with $s_1 \in [0,2]$, the sequence will remain in $[0,2]$, which proves part of the question. Regarding convergence, you can use again the properties of $g$ to check that the sequence in monotone, like you guessed. Note that you can look at $g$ as a real function and use differential calculus...

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  • $\begingroup$ It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet" $\endgroup$ – Wesley Strik Feb 2 at 10:02
  • $\begingroup$ But great answer :) $\endgroup$ – Wesley Strik Feb 2 at 10:03
  • $\begingroup$ Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity. $\endgroup$ – PierreCarre Feb 3 at 12:26
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Wesley's lemma 1 nicely shows that $s_n$ is bounded above.

Rephrasing:

Show that $s_n$ is increasing:

1) $n=1$: $s_2 > s_1√$

2) Hypothesis : $s_{n+1} \ge s_n.$

Step $n+1:$

$s_{n+2} =\sqrt{2+√s_{n+1}} {Hyp. \atop \ge} \sqrt{2+√s_n} =s_{n+1}.$

Used: $f(x)=√x$ is an increasing function.

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    $\begingroup$ You give a nice proof! $\endgroup$ – jackson Feb 2 at 10:24
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    $\begingroup$ Jackson.Thanks, once I got your lead, not so very difficult:)) $\endgroup$ – Peter Szilas Feb 2 at 10:25

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