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Let $\{a_n\}_{n=0}^∞$ be a sequence defined recursively as follows:

$$\left\{\begin{array}{l}a_0=\sqrt2\\a_n=\sqrt{2a_{n-1}},n\geq1\end{array}\right.$$

Show that $\{a_n\}_{n=0}^∞$ converges and that its limit is $2$.

Hint: It may be helpful to first prove that for all $x ∈ (0, 2)$, we have $x < \sqrt{2x} < 2$.


What I tried so far:

I proved the hint first

WTS $$\forall x\in(0,2), x<\sqrt{2x}<2$$

$$\Leftrightarrow \forall x,0<x<2\rightarrow x<\sqrt{2x}\wedge \sqrt{2x}<2$$

Assume the negation:

$$\not\Leftrightarrow \exists x,0<x<2 \wedge (x\geq\sqrt{2x}\vee \sqrt{2x}\geq2)$$

$$\Leftrightarrow \exists x,(0<x<2 \wedge x\geq\sqrt{2x}) \vee (0<x<2 \wedge \sqrt{2x}\geq2)$$

Case1: $\exists x,0<x<2 \wedge x\geq\sqrt{2x}$

Let $x$ be such $x$

Since $0<x<2$

Have $0<\sqrt{2x}$

That $0<\sqrt{2x}\leq x$

$\Rightarrow 0<2x<x^2$

$\Rightarrow 0<2\leq x$ Contradiction

Case2: $\exists x,0<x<2 \wedge \sqrt{2x} \geq 2$

Let x be such x

Have $0<x<2\leq \sqrt{2x}$

$\Rightarrow 0<x^2<4\leq 2x$

$\Rightarrow 0<x<\frac{4}{x}\leq 2$

But $0<x<2$

Implies $4x<8$

$\Rightarrow 4<\frac{8}{x}$

$\Rightarrow \frac{4}{2}<\frac{4}{x}$

$\Rightarrow 2<\frac{4}{x}$ Contradiction

Therefore $\forall x\in(0,2), x<\sqrt{2x}<2$ hold by contradiction.

Then prove $\forall n \in \mathbb{N},0<a_{n}<2$ by induction:

Base case: $n=0$ $$0<\sqrt2<2 \text{ hold}$$

Inductive steps:

Assumse $$0<a_k=\sqrt{2a_{k-1}}<2$$

Show $$0<a_{k+1}=\sqrt{2\sqrt{2a_{k-1}}}<2$$

By assumption $$0<a_k=\sqrt{2a_{k-1}}<2$$

Have $$\sqrt{2(0)}<\sqrt{2(a_k)}=\sqrt{2(\sqrt{2a_{k-1}})}<\sqrt{2(2)}$$

That $$0<a_{k+1}=\sqrt{2\sqrt{2a_{k-1}}}<2 \text{ hold}$$

Therefore $$\forall n \in \mathbb{N},0<a_{n}<2$$

Since $0<a_{n-1}<2$, that $x<\sqrt{2x}<2$ hold for $a_{n - 1}$ implies: $$0 < a_{n - 1} < \sqrt{2a_{n-1}} = a_n < 2,$$ Therefore $a_n$ is increasing and bounded above, implies: $$\exists L\in \mathbb{R},\{a_n\}_{n=0}^∞=L$$

By given also have $$a_n = \sqrt{2a_{n-1}}$$ Take the limit on both sides we have: $$L = \sqrt{2L} \rightarrow L = 2.$$ Therefore $$\{a_n\}_{n=0}^∞=2$$

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  • $\begingroup$ show $a_n \le a_{n+1}$ and $a_n \le 2$ by induction. Then there must be a limit of $(a_n)_n$, call it $L$. $L$ must satisfy $L = \sqrt{2L}$, so $L=2$. $\endgroup$ Jul 28, 2019 at 1:30

4 Answers 4

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First, I just want to say, your proof of the hint is indeed correct, but it's... it's so inelegant. What you have here is the fact that the geometric mean of $x$ and $2$ lies between $x$ and $2$. This can be proven, more directly.

Since $0 < x < 2$, we have $0 < \sqrt{x} < \sqrt{2}$ since the square root function is strictly increasing. Then, multiplying through by $\sqrt{x} > 0$, $$0 < x < \sqrt{2x}.$$ Now, multiplying through by $\sqrt{2} > 0$, $$0 < \sqrt{2x} < 2.$$ In total, $$0 < x < \sqrt{2x} < 2.$$

How does this help? Well, taking $x = a_{n - 1}$ implies $$0 < a_{n - 1} < \sqrt{2a_{n-1}} = a_n < 2,$$ implying both that $a_n$ is increasing, and bounded above. Thus, a limit $L$ exists. If we take the recurrence relation $$a_n = \sqrt{2a_{n-1}}$$ and take the limit of both sides, then $$L = \sqrt{2L} \implies L = 2.$$ The limit must therefore be $2$.

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First we show that the sequence is increasing and bounded above, so it does converge.

Then we show that the limit is in fact $2$.

We know that $a_0 = \sqrt 2 <2 $ and if $a_{n-1}<2$ then $$a_n= \sqrt {2a_{n-1}}<\sqrt {2\times 2} =2$$

In order to show that the sequence is increasing , we have $$\sqrt {2a_{n-1}}>a_{n-1} \iff$$

$$ 2a_{n-1}>a_{n-1}^2 \iff a_{n-1} (2-a_{n-1})>0$$

Since $$0<a_{n-1}<2$$ the above inequality holds therefore the sequence is increasing .

Now that we know the sequence converges to a real number $l$ we take limit of $$ a_n=\sqrt {2a_{n-1}}$$ to get $$l=\sqrt {2l}$$ or $$l^2=2l$$ which gives us $l=0$ or $l=2$

The acceptable limit is $l=2$ since the sequence is increasing and starts at $\sqrt 2$

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  • $\begingroup$ why did you switch from "can show" to "show" in the first sentence? $\endgroup$ Jul 28, 2019 at 3:12
  • $\begingroup$ See if you like the edit. $\endgroup$ Jul 28, 2019 at 3:16
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For the hint, suppose that $0<x<2$, then to the inequality $x<2$ we multiply it by $x$ (which is positive) and we get $x^2<2x$, that is, $x<\sqrt{2x}$. And similarly we obtain $\sqrt{2x}<2$ (multiply the inequality $x<2$ by two). Thus $$x<\sqrt{2x}<2$$ Now, if we know that a sequence is convergent if and only if it is bounded and non-decreasing, it follows from the hint (set $x=a_{n-1}$) that $$a_{n-1}<a_n$$ (it is non-decreasing) and that $$a_n<2$$ for all $n\in \mathbb N$ (the sequence is bounded by $2$).

Finally, to prove that the limit is in fact $ 2 $ just note that, since the sequence is convergent, let's say it converges to $ \ell $. And observe this (after "take the limit" in the recursive formula) : $$\ell =\sqrt{2 \ell}$$ $$\ell (\ell -2)=0$$ and clearly, $\ell$ cannot be equal to $0$ (why?). Therefore $$\lim_{n\to \infty} a_n =2$$

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$$ a_0 = \sqrt 2=2 ^ {1/2} $$ $$ a_1=\sqrt {2\sqrt2} =2 ^{(1/2)+(1/2)^2}$$ $$ ... $$ $$ a_n = 2 ^{(1/2)+(1/2)^2+...+(1/2)^{n+1}} = 2^{\frac{(1/2)^{n+2}-(1/2)}{(1/2)-1}}=2^{1-(1/2)^{n+1}}$$ $$ lim_{n\rightarrow\infty}a_n=lim_{n\rightarrow\infty}2^{1-(1/2)^{n+1}}=2 $$

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