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Find the limit of:

$a_1=\sqrt{3}$

$a_2=\sqrt{3\sqrt{3}}$

$a_3=\sqrt{3\sqrt{3\sqrt{3}}}$

...

By using induction, I found that the limit is 3 but it seems pretty strange to me, I thought it would go to a further number.

I just want to know if I'm right (proof-verification).

puradrogasincortar
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    The limit is $3$, but we can't really verify your proof without more details about your argument. – carmichael561 Sep 11 '17 at 17:57
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    Can't verify a proof if the proof isn't given. Not sure why you think it'd go further. What if I said I was surprised it went that far? – fleablood Sep 11 '17 at 18:06

4 Answers4

3

You can write $a_{n+1} = \sqrt{3a_n}$, so if we mark $ \displaystyle\lim _{n\to \infty}a_n$ we have $$\lim _{n\to \infty}a_{n+1} = \lim _{n\to \infty}\sqrt{3a_n}$$

so $a^2 =3a$ so $a= 0$ or $a=3$. Since it is obviously increasing we have $a=3$.

Add to comment down.

It is obviously that $a_n>0$ for each $n$. Now prove that $a_n<3$ for each $n$ with induction. We do only induction step. So we know $a_n< 3$. Then $a_{n+1}< \sqrt{3\cdot 3} =3$ and we are done. Now prove for increasing

$$a_{n+1}-a_n = \sqrt{3a_n}-a_n = \sqrt{a_n}(\sqrt{3}-\sqrt{a_n}) >0 $$

So sequence is increasing and bounded so it is convergent.

nonuser
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  • Of course before you can do this, you have to show that the limit exists. – GEdgar Sep 11 '17 at 17:59
  • @GEdgar I already did it by prooving its increasing and its bounded, I just wanted to know if the limit was correct! – puradrogasincortar Sep 11 '17 at 18:01
  • " I already did it by prooving its increasing and its bounded" When did you do that? Nowhere do you point out it is bounded. And i don't see why it is obvious it is increasing. – fleablood Sep 11 '17 at 18:07
3

Suggestion:

$$\sqrt{3\sqrt{3\sqrt{3\sqrt{3\cdots}}}} = 3^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots}$$

Victor
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2

The sequence is defined recursively as

$x_0=\sqrt 3;\;x_n=\sqrt{3x_{n-1}}$

It's not difficult to prove that $x_n=3^{1-\frac{1}{2^{n+1}}}$

Indeed if $x=1$ we have $x_1=\sqrt{3\sqrt{3}}=\sqrt[4]{3^3}=3^{\frac34}=3^{1-\frac14}$

If the formula is valid for $n$ then $x_n=3^{1-\frac{1}{2^{n+1}}}$ we show that it is valid for $x_{n+1}=\sqrt{3x_n}=\sqrt{3\cdot 3^{1-\frac{1}{2^{n+1}}}}=\sqrt{3^{2-\frac{1}{2^{n+1}}}}=\left(3^{2-\frac{1}{2^{n+1}}}\right)^{\frac12}=3^{1-\frac{1}{2^{n+2}}}$

Therefore we proved by induction that $x^n=3^{1-\frac{1}{2^{n+1}}}$ and as $n\to\infty$ we have $x\to 3$

More quickly,assuming that the sequence converges, $$x=\sqrt{3\sqrt{3\sqrt{3}}}\ldots$$

$$x^2=3\sqrt{3\sqrt{3}}\ldots$$

$$x^2=3x$$

$$x=3$$

hope it helps

Raffaele
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1)$3 > a_{i+1} > a_i > 1$.

Pf: by induction.

Base step: $a_1 = \sqrt {3} > 1; a_1 = \sqrt{3} < 3$.

Induction: If $3> a_i > 1$ then $a_{i+1} = \sqrt{3*a_i} = \sqrt 3\sqrt{a_i}$

So $3 = \sqrt{3}\sqrt{3} > \sqrt{3}\sqrt{a_i} =a_{i+1}=\sqrt{3}\sqrt{a_i} > \sqrt{3}*1 > 1$.

And $a_{i+1} = \sqrt{3}\sqrt{a_i} > \sqrt{a_i}\sqrt{a_i} = a_i$.

So $3 > a_{i+1} > a+i > 1$.

2) So $a_i$ is increasing and bounded so $\lim\limits_{n\to\infty} a_n$ exists and is less than or equal to $3$.

I honestly have no idea whatsoever why you would have possibly thought the limit would be more than $3$.

Let $\lim\limits_{n\to\infty} a_n = A$

Then $\lim\limits_{n\to \infty} \sqrt{3a_n} = \sqrt{3A}$

$\lim\limits_{n\to\infty}a_{n+1} = \sqrt{3A}$

$\lim\limits_{n+1\to\infty}a_{n+1} = \sqrt{3A}$

$A = \sqrt{3A}$

$A^2 = 3A$

$A =3$ or $A = 0$ but as $A > 1$ we have $A=3$.

fleablood
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