Recently I encountered a strengthened version of this question. Suppose $x_1 > 0$ and $x_{n + 1} = \log(1 + x_n)$. Prove that $$\lim_{n \rightarrow \infty}\frac{n(na_n - 2)}{\log n} = \frac{2}{3}$$
I tried to follow the method for the original question. Define $b_n = 1/a_n$. It appears that $$b_{n + 1} = b_n + \frac{1}{2} - \frac{1}{12b_n} + O\left(\frac{1}{b_n^2}\right)$$ Therefore $$b_n = \frac{n}{2} - \frac{1}{12} \sum_{1 \leq k < n}\frac{1}{b_k} + O\left(\sum_{1 \leq k < n}\frac{1}{b_k^2}\right)$$ Now if we can prove that $$\sum_{1 \leq k < n}\frac{1}{b_k} = 2\log n + O(1)$$ $$\sum_{1 \leq k < n}\frac{1}{b_k^2} = O(1)$$ Then the limit follows easily, since $$\frac{n(na_n - 2)}{\log n} = \frac{n^2 - 2nb_n}{b_n\log n} = \frac{n/3 + O(n/\log n)}{b_n}$$ and we already know that $b_n/n \rightarrow 1/2$. The second asymptotic is already proved here. However I don't know if the first asymptotic is correct or not. We know that $b_n = n/2 + o(n)$. If we can show that the series $$\sum \frac{1}{n + o(n)} - \frac{1}{n}$$ converges, then we can conclude that $\sum 1/b_n = 2\ln n + O(1)$.
We have that
$$o(n)=n\cdot \omega (n) \quad \omega (n)\to 0$$
therefore
$$\frac{1}{n} - \frac{1}{n + o(n)}=\frac{\omega (n)}{n(1+\omega (n))}\sim \frac{\omega (n)}{n}$$
and the given series