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Let, $(x_{n})$ be a sequence define recursively by, $x_{1}=1$ and $x_{n+1}=\frac{1}{2}(x_{n}+\sqrt{3x_{n}})$. Verify that the sequence is bounded above by 4.

By induction we have:

for $n=1$,

$ x_{1}=1<4$.

Now suppose that $x_{n}<4$ is true for some $n \in \mathbb{N}$. Let see if $x_{n+1}<4$ is true:

By definition of $(x_{n})$ we have, $$\frac{1}{2}(x_{n}+\sqrt{3x_{n}})<4$$ $$x_{n}+\sqrt{3x_{n}}<8$$

By the hypothesis, $x_{n}<4$: $$x_{n}+\sqrt{3x_{n}}<4+\sqrt{12}$$

But now, what can I conclued? Can you help me? Thanks.

2 Answers2

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When you write "by definition of $(x_n)$ we have, ...", that is incorrect. That's the desired conclusion, not a fact in your arsenal.

Here's how to fix this. Because $x_n<4$, $3x_n<12<16$. Hence $x_n+\sqrt{3x_n}<4+\sqrt{16}=8$. Hence $\frac{1}{2}(x_n+\sqrt{3x_n})<\frac{1}{2}(8)=4$. Now we use the definition above to conclude $x_{n+1}<4$, as desired.

vadim123
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  • When you chose $16$ it was with the propose of geting $8$? The true is that this sequence converges to $3$, so any number greater or equal to $3$ will bounded the sequence above.Thanks –  Dec 23 '13 at 14:53
  • One more question. Can I use induction to disprove that this sequence is not bounded above by $2$? –  Dec 23 '13 at 15:26
  • The purpose of 16 is to get 8, yes, because 16 has a nice square root. To disprove that it's bounded above by 2, just take successive terms until you find one bigger than 2. – vadim123 Dec 23 '13 at 16:38
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After Induction hypothesis,for $n+1, x_{n+1}= 1/2 \times (x_n +\sqrt{3x_n}) < 1/2\times (4 + \sqrt{3\times4})= 1/2 \times(4+ 2\sqrt{3})= 2+\sqrt{3}<4.$

math
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