Let ${a_1} > 0$ and ${a_n}$ be a sequence so that for any natural number
$a_{n+1} = 1/{a_n} + {a_n}$
Prove that sequence ${a_n}$ is unbounded.
How should I prove that? Can you give me any ideas ...
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1Since $a_{n+1}^2 \geq a_{n}^2+2$, $a_{n}$ grows faster than $\sqrt{2n}$. – Jack D'Aurizio Oct 17 '16 at 18:06
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1Suppose it were bounded, and $B = \sup a_n$. – Daniel Fischer Oct 17 '16 at 18:06
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@JackD'Aurizio I understand the ${a_{n+1}}^2 $ >= ${a_n}^2 + 2$ part, but where did you get this $sqrt(2n)$ to compare it with $a_n$? – Martiiin Oct 19 '16 at 12:20
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@Martiiin: by induction. If $a_{n+1}^2 \geq a_{n}^2+2$, then $a_{n+1}^2\geq (2n+C)$ hence $a_{n+1}\geq \sqrt{2n+C}$. – Jack D'Aurizio Oct 19 '16 at 12:32
1 Answers
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Hint. Show that sequence is positive and increasing. Therefore it has a positive limit. If this limit is finite, say $L$, then the recurrence relation $a_{n+1} = 1/{a_n} + {a_n}$ implies that $$L=\frac{1}{L}+L.$$ What may we conclude?
Robert Z
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@DietrichBurde which is the same as the answer of start wearing purple? – snulty Oct 17 '16 at 18:37
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@Martiiin No. $a_n$ is increasing so it diverges (and it is unbounded) or it has a finite limit $L>0$, such that $L=1/L+L$. Therefore... – Robert Z Oct 19 '16 at 08:07
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.. @RobertZ Therefore, since $ L = 1/L + L $ is not possible, the sequence $a_n$ doesn't have a finite limit L > 0, so it's increasing and it diverges and consequently it is unbounded. – Martiiin Oct 19 '16 at 08:37
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