# Let $a_n$ be a sequence s.t $a_1 > 0 \land a_{n+1} = a_n + \dfrac{1}{a_n}$. Prove that $a_n$ is increasing and tends to infinity

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.

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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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To prove it is an increasing sequence, just notice that the function $f(x)=x+\frac{1}{x}$ is an increasing function for $x>1$. To prove that, just find the first derivative and check if it is positive for $x > 1$

$$f'(x)=1-\frac{1 }{x^2} > 0 \,, \forall x > 1 \,.$$

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-1. This answer is so flawed, on several counts, that it is simply irrelevant. Exercise: Find a function $g$ and a sequence $(b_n)$ such that $b_{n+1}=g(b_n)$ for every $n$, $g$ is increasing and $g(x)\to\infty$ when $x\to\infty$, such that $(b_n)$ converges. – Did Oct 14 '12 at 8:56
Indeed this is wrong. Replace $f$ by $g:x\mapsto x+\frac{1-x}2$ and check the steps of the argument, then the conclusion. – Marc van Leeuwen Oct 14 '12 at 11:24
It was a late night (3 am) when this answer was posted. A mistake can happen. I was looking to evaluate the limit of the sequence and prove that it diverges to $\infty$. So, thanks for your comments. – Mhenni Benghorbal Oct 14 '12 at 11:35
@Serial-downvoters: what's going on? – Mhenni Benghorbal Jan 23 '14 at 3:00

Hint: $$\text{For every} \ \ x>0 \ \ , \ \ x+\frac{1}{x}\geq2$$ increasing with x the value of 1/x will decrease but overall function x+1/x will increase and goes to infinity.

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True. How relevant? – Did Oct 14 '12 at 8:57
I don't know. Yet. Should I remove that? – Salech Alhasov Oct 14 '12 at 9:06
Salech: Who suggested the last edit to your post is a mystery (Community but who is Community?) but this edit is at the same time wrong and misleading. – Did Oct 14 '12 at 13:31
@DonAntonio Please do not approve Edits unless you fully understand the question, the proposed answer and the relevance of the Edit. – Did Oct 14 '12 at 13:38