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I have a sequence: $ a_{n}=\sqrt{3+ \sqrt{3 + ... \sqrt { 3} } } $ , it repeats $n$-times.

and i have to prove that it is a Cauchy's sequence. So i did this: As one theorem says that every convergent sequence is also Cauchy, so i proved that it's bounded between $ \sqrt{3}$ and $ 3 $ (with this one i am not sure, please check if i am right with this one.)And also i proved tat this sequence is monotonic. (with induction i proved this: $ a_{n} \leq a_{n+1} $ so if it's bounded and monotonic, therefore it is convergent and Cauchy. I am just wondering if this already proved it or not? And also if the upper boundary - supremum if you wish - is chosen correctly. I appreciate all the help i get.

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    $\begingroup$ how do you prove that $3$ is an upper bound ? $\endgroup$
    – Surb
    Jun 25, 2015 at 9:42
  • $\begingroup$ @surb 3 is not the upper bound as pointed out in Nemo's answer its about 2.303 which is less than 3 so its bounded below 3. $\endgroup$
    – Warren Hill
    Jun 25, 2015 at 10:18
  • $\begingroup$ I know that it wasn't the most accurate number,but i did not have any problem without being totally specific, so it still works for 3. If it's bounded by 2.303, then 3 is also an upper value...it still proves boundance,although more accurate would be better, but it isn't much of a problem here. $\endgroup$
    – MathIsTheWayOfLife
    Jun 25, 2015 at 10:24
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    $\begingroup$ 3 is not the least upper bound, but it is an upper bound and therefore works just fine. $\endgroup$
    – Klaus Draeger
    Jun 25, 2015 at 11:10
  • $\begingroup$ Yeah, if it wouldn't I would just use another one. $\endgroup$
    – MathIsTheWayOfLife
    Jun 25, 2015 at 11:13

3 Answers 3

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Yes, correct ideas.

For boundedness, you can use induction:

  1. $\sqrt 3<3$, good.
  2. Suppose $a_n<3$ then $a_{n+1}=\sqrt{3+a_n}<\sqrt{3+3}=\sqrt6<3$.
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Once you have boundedness, you can also show the Cauchy property directly. Note that (for nonnegative $u,v$)

$|\sqrt{3+u}-\sqrt{3+v}| = \frac{|u-v|}{\sqrt{3+u}+\sqrt{3+v}} \le \frac{|u-v|}{2\sqrt{3}}$.

So starting from $|a_0-a_n|\le 3-\sqrt{3}$, you get $|a_i-a_j|\le\frac{3-\sqrt{3}}{(2\sqrt{3})^N}$ for all $i,j\ge N$.

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  • $\begingroup$ This works too, but the first thing popped in my head was that convergence and Cauchy equiualence theorem. $\endgroup$
    – MathIsTheWayOfLife
    Jun 25, 2015 at 11:32
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${ a }_{ n+1 }=\sqrt { a_{ n }+3 } $ $\Rightarrow \quad { a^{ 2 } }_{ n+1 }=a_{ n }+3$ as $n\rightarrow \infty $ $\Rightarrow \quad { a^{ 2 } }_{ n+1 }=a_{ n }+3$ $\quad x^{ 2 }=x+3\quad \Rightarrow $ $x^{ 2 }-x-3=0 $ $and\quad it\quad$ convergents to the $x=\frac { 1+\sqrt { 13 } }{ 2 } $

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  • $\begingroup$ I don't need to calculate limit,but thank you. $\endgroup$
    – MathIsTheWayOfLife
    Jun 25, 2015 at 10:20
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    $\begingroup$ This doesn't prove convergence. It derives the limit under the condition it converges. $\endgroup$
    – wlad
    Jun 25, 2015 at 10:32

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