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I’ve already found that the limit $L = 1+\sqrt{3}$ or $L=1-\sqrt{3}$. Since this sequence does not totally an increasing sequence or decreasing sequence, it’s hard for me to except a root. Can anybody give me a hint?

mjw
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scorpy1
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  • I am commenting here on your previous question about maximisation of product knowing the sum, as I cannot do it directly (You have erased it). In the reference I gave https://math.stackexchange.com/q/1774670, I should have explicitly said that you have to look at Dickman's answer : it shows how your isuue is related to number $e$ in particular. – Jean Marie Aug 11 '19 at 09:26
  • @JeanMarie thanks u so much <3 – scorpy1 Aug 11 '19 at 09:36
  • Hint: The sequence of odd-numbered terms is monotonic. So is the even-numbered one. – lhf Aug 11 '19 at 10:53

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If the limit exists, call it $x$.

Then $2x=x^2-2\implies x=\dfrac {2\pm\sqrt{4-4(1)(-2)}}{2(1)}=\dfrac {2\pm2\sqrt3}2=1\pm\sqrt3 $.

Note, $x_n\lt0\,,\forall n\gt1\implies x\le0$. For $\vert x_n\vert\lt1\,,\forall n$.

  • Very appreciate. But Why abs(xn) < 1 in your second line? – scorpy1 Jun 28 '19 at 02:47
  • $\vert x\vert\lt1\implies \frac12x^2\lt1\implies \vert\frac12x^2-1\vert\lt1$. –  Jun 28 '19 at 02:52
  • Yup! Understand! – scorpy1 Jun 28 '19 at 03:08
  • @Chris Custer I'm sorry but your answer isn't rigorous. Two comments : 1) How an you say that $x_n < 0$ for all $n$ ? [First of all it is not true for $n=1$]... 2) How do you know that $(x_n)$ is a convergent sequence ? – Jean Marie Aug 11 '19 at 09:33
  • @JeanMarie You're right. I don't know if it's convergent. It looks like it's negative (after the first term) though. –  Aug 11 '19 at 10:28