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given the inductive sequence $a_1 = 1$; $a_{n+1} = a_n+\frac{1}{a_n}$ I had to prove that $lim_{n\to\infty}a_n \to \infty$.

I proved it by the following way, but I'm not sure I did everything ok so I would appreciate your feedback.

At first, I prove $(a_n)$ is monotonically increasing by induction, in short I assume $a_{n+1} >a_n$ and prove that also $a_{n+2} >a_{n+1}$ as following:

$a_{n+2} = a_{n+1} + \frac{1}{a_{n+1}} = a_{n} + \frac{1}{a_{n}}+\frac{1}{a_{n+1}} \ge a_{n} + \frac{1}{a_{n}} = a_{n+1}$ as needed, thus monotonically increasing.

Secondly, I would like to show that $(a_n)$ converges to +$\infty$. So I have to show that for every $0<M\in R$ exists N such that for every natural n>N, $a_n>M$. =====> I choose N = $a_n$ = M, So I get $a_{n+1} = a_n + \frac{1}{a_n} = M + \frac{1}{M} > M$ as needed.

In conclusion, I get that $(a_n)$ is monotonically increasing and unbounded from above, thus it converges to +$\infty$

by the way I'm 100% sure the thing I did with N=an = M is wrong, or at least semantically is wrong..

Ami Gold
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1 Answers1

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$N=a_n=M$ does not make sense, really. When reading, I have no idea what you did.

You have $M$, and you want to find $N$. The value of $N$ must be defined only in terms of $M$, and if you set $N=M$, you must prove that $a_N>M$, and you did not prove that.

My advice to solving the problem is different:

You already know that $a_n$ is increasing, meaning that it either converges or diverges to $\infty$. Now, assume that it converges, and find a contradiction. Do this by proving that all candidates for a limit are wrong.

Another hint:

Remember, if $a_{n+1} = a_n + \frac1{a_n}$, then what happens if you sent $n\to\infty$ on both sides?

5xum
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  • didn't get quite everything you wrote about proving $a_n>M$ but I do notice that for $n \to \infty$ while supposing that $(a_n)$ tends to infinity, it would be like: L = L+$\frac{1}{L}$. thus L must be $\infty$ – Ami Gold Dec 01 '15 at 17:34
  • @Ami I'm saying that if you set $N=M$, that is not enough. You need to prove that for every $N>M$, you have $a_n>M$, and if you cannot do this, then $N=M$ is not a good choice. The setting of $N=a_n=M$ is wrong on several levels. – 5xum Dec 01 '15 at 17:37
  • @Ami Why must it be $\infty$? – 5xum Dec 01 '15 at 17:38
  • because I get $\frac{1}{L} = 0$, not sure. so in such a case how do I prove that for every N>M $a_n>M$? – Ami Gold Dec 01 '15 at 17:39
  • @Ami You get $\frac1L=0$, and that is a contradiction, because there is no real number such that $\frac1L=0$. Remember, $\infty$ *IS NOT A REAL NUMBER! A limit does not converge* to infinity, it diverges to infinity, and this is just a special way of diverging. You have now proven that the sequence does not converge, and this means it diverges. Together with the fact that it is increasing, this means that it diverges to infinity. – 5xum Dec 01 '15 at 17:41
  • got it. globally, is there another way to prove it using the formal way with the N and the M (as I tried at first)? – Ami Gold Dec 01 '15 at 17:45
  • @Ami Probably, but it would be much more complicated. – 5xum Dec 01 '15 at 18:03
  • This solution is the one I looked for, would appreciate if somebody could help. – Ami Gold Dec 01 '15 at 18:25
  • @Ami But... Why? You have a perfectly good solution, and it may even be the case that there is no nice way to write which $N$ to select when given a value of $M$, and the solution will be much much much uglier and less clear. – 5xum Dec 01 '15 at 19:31