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Well, ive been having a weee bit of problem solving this homework, can anyone give me a hand?

Prove that the sequence $x_{n}$ where $x_{1}=1$ and $x_{n+1} = \sqrt{3x_{n}}$ is convergent and calculate its limit

user111295
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    Note that if the limit $x$ exists, then $x=\sqrt{3\sqrt{3\sqrt{3\dots}}}$. But $x$ is also inside the surd, and we have $x=\sqrt{3x}$, i.e. $x$ is a fixed point for the mapping projecting a term onto its successor. The equation $x=\sqrt{3x}$ is easily seen to have $x=3$ as one solution. That should be a help for constructing a proof. – Jeppe Stig Nielsen Jan 07 '14 at 12:36

6 Answers6

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Hint:

Prove the 3 is an upper bound (by induction:if $x_{n}<3$ then $x_{n+1}<3$)

Prove the whole sequence is strictly increasing (ie. $x_{n+1}>x_{n}$, use the fact that $3$ is an upper bound).

Apply monotone convergence to know that it does converge to some $L$.

Use limit arithmetic to show that $L$ behave the same way the sequence behave, that is $L=\sqrt{3L}$.

Gina
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Hint $x_n=\sqrt{3 \sqrt{3 \ldots\sqrt{3}}} =3^{\frac12+\frac14+\frac18+\cdots+\frac{1}{2^{n}}}$

nadia-liza
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Hints:

1) By induction you can prove that $1\leq x_n\leq 3$.

2) If it converges to $x$ then $x=\lim x_{n+1}=\lim\sqrt{3x_{n}}=\sqrt{3x}$.

drhab
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To prove convergence, prove the sequence is a contraction. See here. To find the limit, solve the equation

$$ x=\sqrt{3x} .$$

Community
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Mhenni Benghorbal
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Hints.

By induction (show the first case): $$\begin{align*}(1)&\;\;x_n\le 3\;\;\forall\,n\in\Bbb N\;:\;x_n:=\sqrt{3x_{n-1}}\stackrel{\text{Ind. Hyp.}}\le\sqrt{3\cdot3}=3\\{}\\(2)&\;\;x_n\le x_{n+_1}\;:\;x_n:=\sqrt{3x_{n-1}}\stackrel{\text{Ind. Hyp.}}\le\sqrt{3x_n}=:x_{n+1} \end{align*}$$

DonAntonio
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Let $y_n=\dfrac{x_n}3$, then

$$y_{n+1}=\sqrt{y_n}$$

so

$$y_{n+1}=\sqrt{y_n}={(y_{n-1})}^{\frac14}=\dotsb=$$

ziang chen
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