Why is $a_{1} > 0 \land a_{n+1}=a_{n}+\frac{1}{a_{n}}$ unbounded?

Question

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?

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. — , Oct 19 '17 at 14:19
@PhysicsMathsLove It is more helpful to provide a link to this excellent formatting guide. — Théophile, Oct 19 '17 at 14:25
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. — 5xum, Oct 19 '17 at 14:26
@Kevin But this only tells us that the sequence does not converge. It might still be bounded, right? — Cm7F7Bb, Oct 19 '17 at 14:34
@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. — , Oct 19 '17 at 15:07

Michael Rozenberg · Accepted Answer · 2017-10-19T14:31:30.303

2

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!

edited Oct 19 '17 at 14:31

answered Oct 19 '17 at 14:23

Michael Rozenberg

194,933

It's a bit complicated including the summation. Why not just use $a_{n+1}^2>a_n^2+2$ – Simply Beautiful Art Oct 19 '17 at 14:56
Because I think it's the same. – Michael Rozenberg Oct 19 '17 at 14:57
Just because two things are the same does not make them the same in simplicity. – Simply Beautiful Art Oct 19 '17 at 16:34

score 1 · Answer 2 · answered Oct 19 '17 at 14:47

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.

score 0 · Answer 3 · answered Oct 19 '17 at 14:37

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.

PhysicsMathsLove · Answer 4 · 2017-10-19T14:47:30.907

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.

Why is $a_{1} > 0 \land a_{n+1}=a_{n}+\frac{1}{a_{n}}$ unbounded?

4 Answers4