Question 44
Let $(a_n)_{n \ge 1}$ be the sequence defined by $a_1 = 1$ and $a_{n+1} = \frac{a_n}{1+\sqrt{1+a_n}}$, for any $n \in \mathbb{N}^*$. Show that $\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \sum_{k=1}^{n} \log_2(1+a_k) = 2$. Traian Tămâian