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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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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.

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).

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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