How prove that $(a_{n})_{n \in \mathbb{N}}$ is convergent with $a_{0}=1$, $a_{n+1}= \sqrt{2a_{n}}$?

Question

Prove that $(a_{n})_{n \in \mathbb{N}}$ is convergent with $a_{0}=1$, $a_{n+1}= \sqrt{2a_{n}}$ and calculate its limit.

You got the info that the square-root-function is strictly monotonic increasing and also continuous.

I'm very confused how this could be solved because there isn't really a sequence given here. It seems to be a recursive one and I also don't know how to form $a_{n+1}$.

Could I form it to $a_{n}+ a_{1}$ ?(I don't care if it's useful here but I'd like to know if this would be correct at first.)

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I have started by forming to $a_{n}$:

$$2a_{n} = (a_{n+1})^{2}$$

$$a_{n}=\frac{(a_{n+1})^{2}}{2}$$

Then I also formed to $a_{1}:$

$$a_{n}+ a_{1} = \sqrt{2a_{n}}$$

$$a_{1} = \sqrt{2a_{n}} - a_{n}$$

But it seems like this won't lead to anything.. Also I don't know what we can do with the given info, $a_{0} = 1$, the continuity and monotonic increasing...

Answer 1

By induction $$a_{n+1}=\sqrt{2a_n}=\sqrt{2\cdot\sqrt{2a_{n-1}} }=2^{1/2}2^{1/4}a_{n-1}^{1/4}=...=a_0\prod_{i=1}^{n+1}2^{\frac{1}{2^{i}}}.$$

As robjohn suggest, define $b_{n+1}=\log(a_{n+1})$ and conclude.

Answer 2

Note that $$ \frac{a_n}{a_{n-1}}=\left(\frac{a_{n-1}}{a_{n-2}}\right)^{1/2}\tag{1} $$ Therefore, by induction $$ \frac{a_n}{a_{n-1}}=\left(\frac{a_1}{a_0}\right)^{1/2^{n-1}}\tag{2} $$ Since $$ \begin{align} a_n &=a_0\prod_{k=1}^n\frac{a_k}{a_{k-1}}\\ &=\prod_{k=1}^n2^{1/2^k}\\ &=2^{\sum\limits_{k=1}^n1/2^k}\tag{3} \end{align} $$ Since $\sum\limits_{k=1}^n1/2^k$ converges, the sequence converges. Since $\sum\limits_{k=1}^\infty1/2^k=1$, this also tells to what the sequence converges.

Answer 3

The most basic, but still readable proof (in my opinion) is by using the following theorem:

A monotonically increasing, bounded sequence of real numbers is convergent.

So we need to prove that $a_n$ is increasing and bounded. Specifically, I will prove that $1\leq a_n \leq a_{n+1} \leq 2$ for all $n$ using induction.

Base case: The inequality clearly holds for $n = 0$, since $1 \leq 1 \leq \sqrt 2 \leq 2$.

Induction step: Assume the inequality holds for $n = k$, i.e. that $1\leq a_k \leq a_{k+1}\leq 2$. We want to show $$ 1 \leq a_{k+1} \leq a_{k+2} \leq 2 $$ The first inequality follows from the induction hypothesis and transitivity of $\leq$ (we assumed $1 \leq a_k$ and $a_k \leq a_{k+1}$). For the second inequality, we have $$ a_{k+1} = \sqrt{a_{k+1}\cdot a_{k+1}} \leq \sqrt{2a_{k+1}} = a_{k+2} $$ where we used $ a_{k+1}\leq 2$, as well as the monotonically increasing nature of the square root.

For the last inequality, we have $$ a_{k+2} = \sqrt{2\cdot a_{k+1}} \leq \sqrt{2\cdot 2} = 2 $$ which concludes the proof that $1 \leq a_{k+1} \leq a_{k+2} \leq 2$.

Answer 4

With $a_0=1$, $a_{n+1}= \sqrt{2a_{n}}$, so show $(a_n)$ is monotone and bounded.

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First show $(a_n)$ is bounded.

Let $P(n)$ be the statement $1\leq a_n\leq 2$.

Base Case

We have $a_0=1$, and $1\leq 1\leq 2$.

Inductive Step

Let $P(k)$ be true: $1\leq a_{k}\leq 2$. Now check validity of $P(k+1)$ from assumed truth of $P(k)$. Since

$$\sqrt{2\cdot1}\leq a_{k+1}= \sqrt{2a_{k}}\leq \sqrt{2\cdot 2}=2$$ we have $1\leq a_{k+1}\leq 2$ as required, and so $(a_n)$ is bounded.

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Now show $(a_n)$ is monotone.

Base Case

We have $a_1=\sqrt{2a_0}=\sqrt{2}\geq a_0=1$. Now $a_2=\sqrt{2a_1}=\sqrt{2\sqrt{2}}$ and so $a_2>a_1$.

Inductive Step

Let $P(k)$ be true: $a_{k}\geq a_{k-1}$. Now check validity of $P(k+1)$ from assumed truth of $P(k)$. Since $$a_k=\sqrt{2a_{k-1}},\quad\text{and}\quad a_{k+1}=\sqrt{2a_k}$$ then, since the square-root-function is strictly monotonic increasing $$a_{k+1}=\sqrt{2a_k}\geq a_k=\sqrt{2a_{k-1}}$$ since we assumed $a_{k}\geq a_{k-1}$.

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Hence the sequence is monotone increasing, bounded above by $2$, and so must converge to some limit $L$, say. Then $$\sqrt{2a_{n}}\to \sqrt{2L}, \qquad\text{as $n\to\infty$}$$ But the sequences $(a_{n})$ and $(a_{n+1})$ converge to the same limit, $$\lim_{n\to\infty}(a_{n+1})=\lim_{n\to\infty}(\sqrt{2a_{n}})$$ from which: $$L=\sqrt{2L}$$ and so $L=2$.

Answer 5

Hint:

Always start by evaluating a few terms.

$$a_0=1=2^0,a_1=\sqrt2=2^{1/2},a_2=\sqrt{2\sqrt2}=2^{1/2+1/4},a_3=\sqrt{2\sqrt{2\sqrt2}}=2^{1/2+1/4+1/8},\cdots$$

As you can show by induction (or by the geometric series formula),

$$a_n=2^{1-1/2^n}=\frac2{\sqrt{\sqrt{\cdots\sqrt2}}}.$$