Limit involving a recursively-defined sequence $$A_{n+1}=A_n+\frac{1}{\sum_{i=1}^n A_i}$$ with $$A_1=1$$
Find out the value of $$\lim_{n→∞}A_n/\sqrt{\log(n)}$$
I used Stolz Theorem, but it seems to be useless.
 A: First, you have that $A_n \ge 1 $ for all $n \ge 1$.
Now let $B_n = {A_n}^2$ and $ C_n=\sum\limits_{k=1}^n A_k $ so you have  $ C_n \ge n$ , $\frac{1}{C_n} \le \frac{1}{n}$, $A_n  =A_1 + \sum\limits_{k=1}^n \frac{1}{C_k} \le 1+\sum\limits_{k=1}^n \frac{1}{k} = 1+H_n = O( \log(n))$ and $C_n = \sum_{k=1}^n A_k =  \sum_{k=1}^n O( \log(k)) = O(n\log(n))$.
Moreover $B_{n+1}={A_{n+1}}^2 = {A_n}^2 + \frac{1}{ {(\sum_1^n A_i)}^2} + 2 \frac{A_n}{(\sum_1^n A_i)} = B_n + \frac{1}{{C_n}^2} + 2\frac{A_n}{C_n}$.
Since $    \frac{1}{{C_n}^2}  =  o\left(\frac{A_n}{C_n}\right)$, $B_{n+1}-B_n \sim 2 \frac{A_n}{C_n}$, therefore $B_n \sim 2 \sum\limits_{k=1}^{n} \frac{A_k}{C_k}$.
Now you have for all $k\ge2$ ,$  \frac{A_k}{C_k}  \le \int\limits_{C_{k-1}}^{C_k} \frac{dt}{t}$ so $\sum_{k=1}^n \frac{A_k}{C_k}  \le \int\limits_{C_{0}}^{C_n} \frac{dt}{t}= \log({C_n})$.
I let you do the other inequality to show that $ \sum_{k=1}^n \frac{A_k}{C_k}  \sim \log({C_n}) \sim \log(n)$.
So finally  $A_n \sim  \sqrt{ 2\log(n)}$ 
