Sequence defined recursively: $\sum_{k=1}^n x_k = \frac{1}{\sqrt{x_{n+1}}}$ Let $(x_n)_{n \ge 1}$ defined as follows:
$$x_1 \gt 0, x_1+x_2+\dots+x_n=\frac {1}{\sqrt {x_{n+1}}}.$$ Compute the limit $\lim _ {n \to \infty} n^2x_n^3.$
MY TRY: I thought about using Stolz-Cesaro lemma, but I couldn't get to an appropriate form that leads to an easier limit.

This answer will use the Stolz-Cesaro lemma.
$$L=\lim_{n \to \infty} n^2x_n^3$$
$$S_n=\sum_{k=1}^n{x_k} \quad\quad x_{n+1}=S_n^{-2}$$
We will try to find a recurrence relation that is easier to manipulate:
$$\frac{x_{n+1}}{x_n}=\frac{S_n^{-2}}{S_{n-1}^{-2}}=\frac{S_{n-1}^{2}}{S_n^{2}}=\left (\frac{S_{n-1}}{S_n}\right)^2$$
$$S_{n-1}=S_n-x_n$$
$$\frac{x_{n+1}}{x_n}=\left(\frac{S_n-x_n}{S_n}\right)^2$$
$$\frac{x_{n+1}}{x_n}=\frac{1}{x_nS_n^2}=\left(\frac{1}{\sqrt{x_n}S_n}\right)^2$$
$$\left(\frac{S_n-x_n}{S_n}\right)^2=\left(\frac{1}{\sqrt{x_n}S_n}\right)^2$$
$$\frac{S_n-x_n}{S_n}=\frac{1}{\sqrt{x_n}S_n}$$
$$S_n-x_n=\frac{1}{\sqrt{x_n}}$$
$$S_n=x_n+\frac{1}{\sqrt{x_n}}=x_n+x_n^{-\frac{1}{2}}$$
$$x_{n+1}=\frac{1}{\left(x_n+x_n^{-\frac{1}{2}}\right)^2}$$
$$x_{n+1}=\frac{x_n}{\left(x_n^{\frac{3}{2}}+1\right)^2} \quad n>1$$
Because of the limit we want to evaluate, we make the substitution: $a_n=n^2x_n^3 \quad x_n=a_n^{\frac{1}{3}}n^{-\frac{2}{3}} \quad L=a_{\infty}$
$$a_{n+1}^{\frac{1}{3}}(n+1)^{-\frac{2}{3}}=\frac{a_n^{\frac{1}{3}}n^{-\frac{2}{3}}}{\left(\frac {\sqrt{a_n}}{n}+1\right)^2}$$
$$a_{n+1}(n+1)^{-2}=\frac{a_nn^{-2}}{\left(\frac {\sqrt{a_n}}{n}+1\right)^6}$$
$$a_{n+1}=\frac{a_n(n+1)^{2}}{n^2\left(\frac {\sqrt{a_n}}{n}+1\right)^6}$$
To get rid of the square root we make the substitution: $a_n=b_n^2 \quad b_n=\sqrt{a_n} \quad L=b_{\infty}^2$
$$b_{n+1}^2=\frac{b_n^2(n+1)^{2}}{n^2\left(\frac {b_n}{n}+1\right)^6}$$
$$b_{n+1}=\frac{b_n(n+1)}{n\left(\frac {b_n}{n}+1\right)^3}$$
$$\frac{b_{n+1}}{n+1}=\frac{b_n}{n\left(\frac {b_n}{n}+1\right)^3}$$
This next substitution is somewhat obvious: $c_n=\frac{b_n}{n} \quad b_n= nc_n \quad L=\lim_{n \to \infty} (nc_n)^2$
$$c_{n+1}=\frac{c_n}{\left(c_n+1\right)^3}$$
Because we want the limit in a usable form for the Stolz-Cesaro lemma, we make the substitution: $c_n=\frac{1}{d_n} \quad d_n=\frac{1}{c_n}$
$$L=\lim_{n \to \infty} \left(\frac{n}{d_n}\right)^2=\left( \lim_{n \to \infty} \frac{n}{d_n}\right)^2$$
$$L=\left( \lim_{n \to \infty} \frac{n+1-n}{d_{n+1}-d_n}\right)^2$$
$$L=\lim_{n \to \infty} \frac{1}{(d_{n+1}-d_n)^2}$$
$$\frac{1}{d_{n+1}}=\frac{1}{d_n\left(\frac{1}{d_n}+1\right)^3}$$
$$d_{n+1}=d_n\left(\frac{1}{d_n}+1\right)^3=\frac{(d_n+1)^3}{d_n^2}$$
$$d_{n+1}=\frac{d_n^3+3d_n^2+3d_n+1}{d_n^2}$$
$$d_{n+1}=d_n+3+\frac{3}{d_n}+\frac{1}{d_n^2}$$
$$\lim_{n \to \infty} \left(d_{n+1}=d_n+3+\frac{3}{d_n}+\frac{1}{d_n^2}\right)$$
$$\lim_{n \to \infty} \left(d_{n+1}=d_n+3\right)$$
$$\lim_{n \to \infty} d_{n+1}-d_n=3$$
$$L=\frac{1}{3^2}=\frac{1}{9}$$
All of this is assuming that $d_n$ diverges to $+\infty$. Since $x_1>0$, $d_1>0$ (substitutions considered) is the only case that needs to be analyzed. We will prove that $d_{n+1}>d_n$ for $d_n>0$.
$$\frac{d_n^3+3d_n^2+3d_n+1}{d_n^2}>d_n$$
$$d_n^3+3d_n^2+3d_n+1>d_n^3$$
$$3d_n^2+3d_n+1>0$$
This is clearly true for $d_n>0$. Note that this also shows there are no solutions to $d_{n+1}=d_n$ for $d_n>0$. Therefore, we can conclude $L=\frac{1}{9}$
