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$\lim_{x\to\infty} a_n$ defined as $a_{n+1}=a_n+\frac{1}{a_n}$ with $a_1=1$.

I've tried writing out the first few terms and this sum is increasing, but progressively less-so over time. I'm just not certain how to go about caclulating the limit here. I feel that it does have a finite limit, but I'm not certain of what it is.

MC989
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3 Answers3

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It can't converge. Otherwise, if $\lim_{n\to\infty} a_n=L$:

$$\lim_{n\to\infty} a_{n+1}=\lim_{n\to\infty} a_n+\frac{1}{a_n}$$ $$L=L+\frac1L$$ $$0=\frac1L$$

Which is a contradiction.

  • it can converge to $\infty$.. your statment is false. – Mykybo Dec 05 '15 at 20:00
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    @Mykybo Most of the time, we call that "diverging", unless we are working in nonstandard analysis which was not stated. –  Dec 05 '15 at 20:01
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HINT:

$$a_{n+1}^2 = a_n^2 + 2 + \frac{1}{a_n^2} > a_n^2 + 2$$

so $(a_n^2)_n$ grows at least as fast as an arithmetic progression (say $a_n^2 > 2n-1$ for $n>1$).

orangeskid
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  • If one had a recurrence like $a_{n+1} = a_n + \frac{1}{a_n^2}$ the trick would be to consider $a_{n+1}^3$ ... – orangeskid Dec 05 '15 at 21:00
  • see math.stackexchange.com/questions/1540572/upper-bound-of-a-n1-a-n-frac1a-n where they start with $a_0$ rather than $a_1.$ Just need to shift the index $n$ by $1$ to match them up. – Will Jagy Dec 05 '15 at 21:05
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    @Will Jagy: Yes, that appears to be a finer thing, I'll think about it, thanks! – orangeskid Dec 05 '15 at 21:38
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    I put in my own answer rather late; it was the material for something more extreme, I will make something up, $ a_0 > 0$ and $ a_{n+1} = a_n + \frac{1}{e^{a_n}}.$ There may be no elegant approach, but you can still conclude that the sequence is unbounded and give a lower bound in terms of $n$ – Will Jagy Dec 05 '15 at 22:34
  • @Will Jagy: Indeed, any sequence satisfying $a_{n+1} = a_n + f(a_n)$, with $f$ (continuous and) $>0$ on $(0, \infty)$ has to be unbounded. I wonder what effective lower bounds can one find for your example above. – orangeskid Dec 05 '15 at 23:15
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    Oh, exactly what I put in my answer http://math.stackexchange.com/questions/1540572/upper-bound-of-a-n1-a-n-frac1a-n/1540580#1540580 setting my $f(x) = e^{-x},$ so that my $N_j = \left\lceil e^j \right\rceil$ – Will Jagy Dec 05 '15 at 23:21
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    @Will Jagy: Oh, that is very nice and clever! – orangeskid Dec 05 '15 at 23:25
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Outline:

  • Show that $(a_n)_n$ is increasing (by induction: note that $a_{n+1} - a_n = \frac{1}{a_n} \geq 0$, as you can (and will) show that $a_n \geq a_1 = 1$)
  • show that it converges to $\infty$, by contradiction: if it had a limit $\ell < \infty$, then what equation would $\ell$ satisfy? (note that since the sequence is monotonically increasing, it can only have a limit or diverge to $\infty$)
Clement C.
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  • Now how do I use induction to show increasing? I started with $n=2$ as a base case, then assumed my IH to be $a_{n+1}>a_n$ and applied to $a_{n+2}=a_{n+1}+\frac{1}{a_{n+1}}$ but how do I proceed? Even if I break it down into terms of $a_n$, how do I know this is increasing? – MC989 Dec 05 '15 at 20:55
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    $a_2 = a_1+\frac{1}{a_1} = 2 > a_1$. Then, assuming the property up to $n$, $$a_{n+1} - a_n = \frac{1}{a_n} > 0$$ using the fact that $a_n \geq a_1 = 1 > 0$. So $a_{n+1} > a_n$. – Clement C. Dec 05 '15 at 20:57
  • Do you think I can assume that from the given problem that ${a_n}$ is unbounded, thus since it is an increasing and unbounded sequence, its limit is $\infty$? I have a theorem in my book that says as much, I'm just not certain about the unbounded part. – MC989 Dec 05 '15 at 21:15
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    Did you read the hint in my post? It's the same idea as in avid19's answer: if there is a finite limit $\ell$, then by taking the limit in the recurrence relation (which is legit, as $a_{n+1}$ a continuous function of $a_n$), then $\ell = \ell+\frac{1}{\ell}$. Besides that approach, to claim the sequence is unbounded you'd have to prove it... – Clement C. Dec 05 '15 at 21:17