# Equivalent of a divergent sequence

I've read in an article of this site that if you consider the sequence defined for $n \in \mathbb{N}$ by $$u_0=0 \text{ and }u_{n+1}=\left(u_n\right)^2+\frac{1}{4}$$ Then the sequence $\displaystyle \left(u_n\right)_{n \in \mathbb{N}}$ converges to $1/2$, no problem so far. However, it also explains that the sequence for $\alpha>0$ given by $$v_0=0 \text{ and }v_{n+1}=\left(v_n\right)^2+\frac{1}{4}+\alpha$$ is a divergent sequence but if we let $N$ be the number of iteration needed so that $v_N>2$,it was written that $$N\left(\alpha\right)\underset{(0^{+})}{\sim}\frac{\pi}{\sqrt{\alpha}}$$ Is there a way to show this equivalence ? Here is a link to the article

• "I've read in an article of this site" – what about adding a link to that article, so that we know the context? – Martin R Apr 10 '18 at 13:01
• The article can be found at pdfs.semanticscholar.org/dbed/…. – Ewan Delanoy Apr 10 '18 at 14:38
• Isn' there already a proof in the article? – Seewoo Lee Apr 12 '18 at 7:15