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Let $u_0=1$ and $u_{n+1}=\frac{u_n^2}{u_n+1}, \forall n\in \mathbb{N}$.

a) Find the formula of $u_n$?

b) Calculate the limit $\displaystyle\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}$.

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In a), what is the meaning of the formula? Note that $x_n=1/u_n$ is such that $x_{n+1}=x_n^2+x_n$ and that quadratic functions are notoriously difficult to iterate, as witnessed by the Julia and Mandelbrot sets. –  Did Feb 29 '12 at 17:40
    
For b), note that $u_{n}\leqslant2^{-2^{n-1}}$ for every $n\geqslant1$. –  Did Feb 29 '12 at 17:41
    
It helps to look at $v_n = \frac{1}{u_n}$ instead. Then the recurrence becomes $v_0 = 1$ and $v_{n+1} = v_n (v_n + 1)$ which is easier to understand. –  TMM Feb 29 '12 at 18:43
    
That $v_{n+1}=v_n(v_n+1)$ is wrong. It is $v_{n+1}=v_n+1/v_n$ –  GEdgar Feb 29 '12 at 22:33
    
@GEdgar: Really? –  Did Mar 2 '12 at 18:49
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3 Answers

up vote 1 down vote accepted

For a "formula"

Using the methods in the paper: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf, and as mentioned in the comments to the OEIS sequence: https://oeis.org/A007018 I believe we can show that

$$\frac{1}{u_n} = \text{largest even integer less than } k^{2^n}$$ for $n \ge 1$.

where $k$ is a constant (definable as a limit of a sequence defined in terms of $\{u_n\}$, see that paper).

Basically, the methods in the paper can be used to show that

$$\frac{1}{u_n} \lt k^{2^n} \lt \frac{1}{u_n}+2$$

Since $\frac{1}{u_n}$ is an even integer for $n\ge 1$, the result follows.

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Hint for b):

Show that $u_n\rightarrow0$ (it's easy to show that the sequence is positive, bounded below by 0, and is decreasing. So, it converges and its limit satisfies the formula $L=L^2/(L+1)$).

Then use the fact that, for a sequence $(a_n)$ of positive numbers, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n} $exists, then $\lim\limits_{n\rightarrow\infty} \root n\of{a_n}$ exists and the two limits are equal.

This may be a "sledgehammer", though...




Incidentally: The first few terms of the sequence sequence are $1, {1\over2}, {1\over2\cdot3}, {1\over 6\cdot 7}, {1\over 42\cdot43}, {1\over 1806\cdot 1807}, \ldots$. The denominators listed here, when entered into OEIS, gives an interesting result (see sequence A100016 at the bottom).

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It's only seen by some as a sledgehammer because the theorem about the two limits is relatively unknown. I don't know why many people don't know it, considering people know of the Root and Ratio tests and this theorem is quite similar material. +1 by the way. –  Ragib Zaman Feb 29 '12 at 17:25
    
Dear David Mitra. Please give me a proof of your lats statement. If $\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ exists, then $\displaystyle\lim_{n\rightarrow\infty}a_n^{\frac{1}{n}}$ exists and two limits are equal. –  impartialmale Mar 17 '12 at 21:53
    
@impartialmale See Lemma 3 in these notes of Pete L. Clark's. I think the proof is also in Rudin's (baby) analysis text. You should be able to find a proof in most Real Analysis texts. –  David Mitra Mar 17 '12 at 22:12
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Taking Didier's comment, write in fact $z_n = (1/u_n)+(1/2)$ to get recurrence $$ z_{n+1} = z_n^2+\frac{1}{4} $$ and then consult the literature about the Mandelbrot set. It is known that the recurrence $z_{n+1} = z_n^2+c$ has "closed form" (in a certain precise sense$^1$) if and only if $c=0$ or $c=-2$. Thus, in this case $c=1/4$ it has no "closed form".

$^1$Eremenko, page 663, in: L. Rubel, "Some research problems about algebraic differential equations, II" Ill. J. Math. 36 (1992) 659--680.

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