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This question already has an answer here:

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?

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marked as duplicate by Arnaud D., Christopher, Paul Frost, Wouter, Vidyanshu Mishra Oct 5 '18 at 13:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @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. $\endgroup$ – Paul Frost Oct 5 '18 at 12:25
  • $\begingroup$ @PaulFrost This meta question already exist in fact : math.meta.stackexchange.com/questions/16417/… (But I admit that this case is far from clear-cut.) $\endgroup$ – Arnaud D. Oct 5 '18 at 12:34
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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 $a_n \to \infty$.

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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.

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