Let $(x_n)_{n \in \mathbb{N}}$ be a sequence such that $x_{n+1} = \sqrt{3+2x_n}$ and $x_0 \in [0, 3]$. Prove that: $$\lim_{n \to \infty} n^k(3 - x_n) = 0, \, \, \forall \, k \in \mathbb{N}^*$$

I've proved that $(x_n)$ is increasing, bounded and has a limit of $3$, but I haven't managed to prove that the given limit is $0$.

Thank you in advance!


Let $x_n=3-y_n$ and the recursion becomes

$$3-y_{n+1} = \sqrt{9-2 y_n} = 3 \sqrt{1-\frac{2}{9} y_n} \approx 3-\frac13 y_n$$

i.e., for large $n$, $y_{n+1} \approx y_n/3$, or $y_n \approx c \cdot 3^{-n}$, which indeed disappears faster than any power of $n$.


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