limits - 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.

Answer

Let $y_n=x_1+\ldots+x_n$ for $n\ge1$. Then, the recursion reads $y^2_n\,(y_{n+1}-y_n)=1.$ We see that $y_{n+1}-y_n>0$ for $n\ge1.$ Since $y^2$ is monotone, it follows that
$$n=\sum^n_{k=1}y^2_k\,(y_{k+1}-y_k)\le\int^{y_{n+1}}_{y_1}y^2\,dy=\frac13(y^3_{n+1}-y^3_1),$$
i.e. $$y_{n+1}\ge3^{1/3}n^{1/3}.\tag1$$ Consequently $$\frac1{y^2_n}\le3^{-2/3}\frac1{(n-1)^{2/3}},$$ and $$y_{n+1}=y_2+\sum^n_{k=2}\frac1{y^2_k}\le y_2+3^{-2/3}\sum^n_{k=2}\frac1{(k-1)^{2/3}}=y_2+3^{-2/3}\sum^{n-1}_{k=1}\frac1{k^{2/3}}.$$ Now we can estimate the RHS with a telescope: $$k^{1/3}-(k-1)^{1/3}=\frac1{k^{2/3}+k^{1/3}(k-1)^{1/3}+(k-1)^{2/3}}\ge\frac1{3k^{2/3}},$$ that means $$y_{n+1}\le y_2+3^{1/3}(n-1)^{1/3}.\tag2$$
(1) and (2) together give $$\lim_{n\rightarrow\infty}y_n\,n^{-1/3}=3^{1/3}.$$

Since $x_{n+1}=y_{n+1}-y_n=1/y^2_n,$ we have $$\lim_{n\rightarrow\infty}x_n\,n^{2/3}=3^{-2/3},$$ and thus $$\lim_{n\rightarrow\infty}x^3_n\,n^2=3^{-2}=\frac19.$$