Given a recurrence formula, evaluate $\lim\limits_{n\to \infty} n^2 x_n^3$ 
Define a sequence $(x_n)_{n\geq 0}$ with a fixed initial term $x_0 > 0$ such that:
$$x_0 + x_1+\ldots+x_n=\frac{1}{\sqrt{x_{n+1}}}$$
Evaluate 
$$\lim_{n\to \infty} n^2 x_{n}^3$$

My attempt: I should define a new sequence $s_n = \displaystyle\sum_{i = 0}^nx_i$ with the recurrence formula:
$$s_{n+1} = s_n+\frac{1}{s_n^2}$$
$(s_n)_{n\geq 0}$ is increasing and divergent, and I want the limit of:
$$n^2x_n^3 = \frac{n^2}{s_{n-1}^6}$$
So, I can look instead for the limit:
$$\lim_{n\to \infty} \frac{s^3_n}{n}$$
For $s_n^3$, the recurrence formula gives
$$s_{n+1}^3 = \left(\frac{1+s_n^3}{s_n^2}\right)^3$$
Looking at the function $f:(0, \infty) \to (0, \infty),\ f(x) = \dfrac{(x+1)^3}{x^2}$ this is increasing from a certain point forward and it feels that I should squeeze it between $3n$ and $3n+\text{something negligible}$, but I can't give a solid argument for this.

Start with
$$
x_n=\left(\sum_{k=0}^{n-1}x_k\right)^{-2}\tag1
$$
Let $x_n=u_n^{-2}$, then we get
$$
u_n=\sum_{k=0}^{n-1}u_k^{-2}\tag2
$$
From $(2)$, we get
$$
u_{n+1}-u_n=u_n^{-2}\tag3
$$
Equation $(3)$ indicates that $u_n$ is increasing. If $u_n$ were bounded above, it would approach a finite limit, $\bar u$, and then we would have $\bar u=\bar u+\bar u^{-2}$, which is impossible. Thus, $u_n\to\infty$.
Furthermore,
$$
\begin{align}
u_{n+1}^3-u_n^3
&=\left(u_{n+1}^2+u_{n+1}u_n+u_n^2\right)\left(u_{n+1}-u_n\right)\\
&=\left(\frac{u_{n+1}}{u_n}\right)^2+\frac{u_{n+1}}{u_n}+1\\
&=\left(1+u_n^{-3}\right)^2+\left(1+u_n^{-3}\right)+1\tag4
\end{align}
$$
Equation $(4)$ and Stolz–Cesàro then says that
$$
\lim_{n\to\infty}\frac{u_n^3}{n}=3\tag5
$$
which is equivalent to
$$
\lim_{n\to\infty}n^2x_n^3=\frac19\tag6
$$
