Prove a recursive relation

Question

Suppose one has the following recursive relation :

$$ a_{n+1} = a_n + \dfrac{1}{a_n} $$ Where: $$a_n > 0$$

Is there any way to find a closed form formula for something like this? I tried looking at the local factors by looking at how it develops, but it seems very chaotic. I found an inverse formula for this relation:

$$ a_{n-1} = \dfrac{a_n+ \sqrt{{a_n}^2 - 4}}{2} $$

A bit odd is the fact that the inverse function is undefined ( in the reals) for $a_n < 2$ even though the regular function is defined for values lower than 2. Also, since the original function is increasing and monotonic, we know that the inverse is decreasing and monotonic - which means that for very large values of $a_0$ we always expect that the limit as n goes to infinity to be undefined, as it is guaranteed to go lower than 2 at some point in the sequence.

Thats all I got basically, Would love to hear if you have any ideas or know anything about these types of recursive relations, because I noticed for example:

$$ a_{n+1} = a_n + \dfrac{1}{2a_n} $$ has a very similar inverse: $$ a_{n-1} = \dfrac{a_n+ \sqrt{{a_n}^2 - 2}}{2} $$

@lhf find the nth term in the sequence without computing all of the middle terms — dvd280, Aug 10 '20 at 14:00
Related https://math.stackexchange.com/questions/3767681/if-a-1-1-and-for-n1-a-n-a-n-1-frac1a-n-1-then-a-246-lie — Anindya Prithvi, Aug 10 '20 at 14:38
@AnindyaPrithvi not a duplicate, the other one deals with a specific case of a_1 = 1 — dvd280, Aug 10 '20 at 14:39

Robert Z · Accepted Answer · 2020-08-10T14:34:43.657

3

I can't provide a closed formula, but here is double inequality which show how fast $(a_n)_n$ goes to infinity.

Let $a_0>0$ and let $b_n=a_n^2-2n$. Then $b_0=a_0^2>0$ and $$b_{n+1}=\left(a_n+\frac{1}{a_n}\right)^2-2n-2=a_n^2+\frac{1}{a_n^2}-2n=b_n+\frac{1}{b_n+2n}$$ which implies by induction that $b_n>0$ for all $n\geq 0$. Moreover $$b_{n+1}-b_n<\frac{1}{2n}\implies b_{n+1}-b_1<\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\leq \frac{\ln(n)+1}{2}$$ and therefore, since $b_1=a_0^2+\frac{1}{a_0^2}$, we find that $$\sqrt{2n}<a_n<\sqrt{2n+a_0^2+\frac{1}{a_0^2}+\frac{\ln(n)+1}{2}}.$$

edited Aug 10 '20 at 14:34

answered Aug 10 '20 at 14:28

Robert Z

145,942

this is interesting, since we know $a_n > 1 $ it means that the inequality is preserved when raising to the second power. doing this gives $ 2n< {a_n}^2 < 2n + b_1+\dfrac{ln(n)+1}{2}$ – dvd280 Aug 10 '20 at 14:36
1

Yes, it follows from the definition of $b_n$. – Robert Z Aug 10 '20 at 14:37

Prove a recursive relation

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