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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 increasing and tends to infinity.

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.

Now this is where I need some help. I would like to say that $a_n$ is unbounded and then conclude that monotone and unbounded implies tending to infinity.

Maybe by contradiction?

Martin Sleziak
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CodeKingPlusPlus
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  • @ArnaudD. I agree. But an interesting "meta question" is this: Should older questions be closed as duplicates of a more recent questions? Perhaps it depends on the quality of the answers. – Paul Frost Oct 05 '18 at 12:25
  • @PaulFrost This meta question already exist in fact : https://math.meta.stackexchange.com/questions/16417/original-post-marked-as-duplicate (But I admit that this case is far from clear-cut.) – Arnaud D. Oct 05 '18 at 12:34

2 Answers2

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Yes, contradiction will work. Assume the sequence is bounded by $M$. Then $a_{n+1}=a_n+1/a_n> a_n+1/M$. By induction, we have that $a_{n+k}>a_N+k/M$. Thererefore $a_{n+M^2}>a_n+M^2/M=a_n+M>M$, which is a contradiction.

Aaron
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You 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 and it follows that $a_n \to \infty$.

Beni Bogosel
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