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.
 A: You had the right idea of using the series of $(x_n)_n$, and you are almost done.
Observe that $x_n > 0$ for every positive integer $n$, from the definition of the recurrence relation. Also, since \begin{equation} x_{n+1} = s_{n+2} - s_{n+1} = \frac{1}{\sqrt{x_{n+2}}} 
 - \frac{1}{\sqrt{x_{n+1}}} > 0 \end{equation} you get that $x_n \geq x_{n+1}$, so the sequence $(x_n)_n$ is decreasing. As it is also bounded by $x_0$ and $0$, it is convergent, with some limit $l \in \mathbb{R}, l \geq 0$. If $l \neq 0$, then letting $n \to \infty$ in the above equation, you would get that $l = 0$, which is a contradiction. Hence $\displaystyle\lim_{n \to \infty} (x_n) = 0.$ This also implies that $(s_n)_n$ is increasing and tends to infinity.
Now, use the Cesaro-Stolz theorem: $$\lim_{n \to \infty} \left(\frac{n}{s_{n-1}^3} \right) = \lim_{n \to \infty} \left(\frac{1}{s_n^3 - s_{n-1}^3}\right) = \lim_{n \to \infty} \left(\frac{1}{\left( \frac{1}{\sqrt{x_{n+1}}} - \frac{1}{\sqrt{x_n}} \right) \cdot \left( \frac{1}{(\sqrt{x_{n+1}})^2} + \frac{1}{\sqrt{x_{n+1} \cdot x_n}} + \frac{1}{(\sqrt{x_n})^2} \right)} \right).$$
Now, since, $$\frac{1}{\sqrt{x_{n+1}}} - \frac{1}{\sqrt{x_n}} = x_n, \forall n, $$ let $$L(x_n, x_{n+1}) = x_n \cdot \left( \frac{1}{(\sqrt{x_{n+1}})^2} + \frac{1}{\sqrt{x_{n+1} \cdot x_n}} + \frac{1}{(\sqrt{x_n})^2} \right). $$ Observe that $$L(x_{n}, x_{n+1}) \leq \frac{3x_n}{x_{n+1}} $$ and that $$L(x_n, x_{n+1}) \geq \frac{3x_{n+1}}{x_n} $$ since $x_n \geq x_{n+1}$ and since $\displaystyle \frac{1}{\sqrt{x_n}} \leq \frac{1}{\sqrt{x_{n+1}}}$, for all positive integers $n$, and now you only need to prove that $\displaystyle \frac{x_n}{x_{n+1}} \to 1$ as $n \to \infty$.
However, observe that $$\frac{\frac{1}{\sqrt{x_{n+2}}}}{\frac{1}{\sqrt{x_{n+1}}}} = \frac{s_{n+1}}{s_{n}} = 1 + \frac{x_{n+1}}{s_n} \to 1 \text{ as } n \to \infty, $$ so $L(x_n, x_{n+1}) \to 3$ as $n \to \infty$.
A: 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
$$
\lim_{n\to\infty}n^2x_n^3=\frac19\tag6
$$
