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Let $(a_n)$ be a sequence s.t $$a_{1} > 0 \land a_{n+1}=a_{n}+\frac{1}{a_{n}}$$

Prove that $a_{n}$ is unbounded.

Proof:

Consider $a_{n+1}−a_{n}$:

$a_{n+1} - a_{n} = a_{n} + \frac{1}{a_{n}} - a_{n} = \frac{1}{a_{n}}$.

This is greater than $0$. Thus, $a_{n}$ is increasing.

It was proved that $a_{n}$ is increasing. Assume that it is bounded. Then it would follow that $a_{n}$ is convergent to a real number $L>0$. But taking $n\to\infty$ into the recurrence relation gives $$L+\frac{1}{L} =L$$ which is a contradiction. Therefore $a_{n}$ is unbounded

I found this on the site but I don't get why it is unbounded. Could someone plz explain?

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    $\begingroup$ Please use Latex. $\endgroup$ Oct 19, 2017 at 14:18
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    $\begingroup$ Assume it converges, then it does so to a limit $L \geq 1$ Then we have $$L=L+\frac{1}{L}$$ which is not possible. $\endgroup$
    – user284001
    Oct 19, 2017 at 14:19
  • $\begingroup$ @PhysicsMathsLove It is more helpful to provide a link to this excellent formatting guide. $\endgroup$
    – Théophile
    Oct 19, 2017 at 14:25
  • $\begingroup$ If you assume the sequence is bounded, then you arrive at a contradiction. Reductio ad absurdum then tells you that means the sequence is unbounded. $\endgroup$
    – 5xum
    Oct 19, 2017 at 14:26
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    $\begingroup$ @Cm7F7Bb $a_{n+1}-a_n > 0$ and as such $a_n$ is a strictly increasing sequence that does not converge (by dint of my previous comment). Therefore it must be unbounded. $\endgroup$
    – user284001
    Oct 19, 2017 at 15:07

4 Answers 4

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In your case $a$ is unbounded because if $a$ is bounded then there is a limit of $a$.

It gives a contradiction.

Thus, the assuming was wrong, which says that $a$ is unbounded.

Also we can use the following reasoning.

Since $$a_{n+1}^2=a_n^2+2+\frac{1}{a_n^2},$$ For all $n\geq2$ we obtain $$a_n^2=2(n-1)+a_1^2+\sum_{k=1}^{n-1}\frac{1}{a_k^2}>2(n-1),$$ which gives $a_n>\sqrt{2(n-1)}$ and we are done!

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  • $\begingroup$ It's a bit complicated including the summation. Why not just use $a_{n+1}^2>a_n^2+2$ $\endgroup$ Oct 19, 2017 at 14:56
  • $\begingroup$ Because I think it's the same. $\endgroup$ Oct 19, 2017 at 14:57
  • $\begingroup$ Just because two things are the same does not make them the same in simplicity. $\endgroup$ Oct 19, 2017 at 16:34
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Since the equation $x= x+\frac1x$ does not have a solution. Therefore $a_n$ does not converges. Also, since $a_1>0,$ by induction we easily have $a_n>0$ and then, $$a_{n+1} -a_n = \frac{1}{a_n}>0$$ which means $(a_n)$ is a strictly increasing and non convergent sequence. So $a_n\to\infty$. That $a_n$ is unbounded.

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Assume that the sequence converges.

From Cauchy criterion, $\{a_n\}$ converges when, for all $\epsilon>0$, there is a fixed number $N$ such that $|a_j-a_i|<\epsilon$ for all $i,j>N$.

Fix $j=n+1$, $i=n$. Then $$\left|a_{n+1}-a_{n}\right|<\epsilon\quad \Rightarrow \quad \left|\frac{1}{a_{n}}\right|<\epsilon \quad \Rightarrow \quad |a_n|>\frac{1}{\epsilon}\, ,$$ which is a contradiction.

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Note that $$a_{n+1} - a_n = a_n +\frac{1}{a_n} - a_n = \frac{1}{a_n}>0$$ since $(a_n) > 0 $ for all $n$. Therefore the sequence is strictly increasing.

So we either have:

  • the sequence converges and is bounded (can you prove this?)
  • the sequence does not converge and is unbounded (can you prove this?)

Assume it does converge. Then, we know that the limit $l$ must be greater than 1 or equal to 1. By the shift rule, if the limit exists, it satisfies $$l= l+\frac{1}{l}$$ which is impossible. Then the sequence does not converge.

Therefore, we have a strictly increasing sequence that does not converge. It must therefore be unbounded.

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