With $x_1=1$ and $x_{n+1} = \frac{1}{x_1^2+x_2^2+\dots+x_{n}^2}$show$\lim\limits_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n^{2/3}}=\frac{\sqrt[3]{9}}{2}$

Question

Context

I found an extra-hard (but fake) example of a paper which adheres to the syllabus for the hardest mathematics high school course available in the state of NSW in Australia; Mathematics Extension 2. Question 16 (a) has caught my eye and I'm having difficulty solving it, so am looking for tips. I will reproduce the question entirely below and add explain where I am up to.

Question Statement

$16 (a)$ $[\mathbf7$ $\mathbf{marks}]$ A sequence $x_1, x_2, x_3, \dots$ of positive real numbers is given, recursively defined by $x_1=1$ and $x_{n+1} = \frac{1}{x_1^2 + x_2^2 + \dots + x_{n}^2}$ for each $n > 1$.

$(i)$ $[\mathbf2]$ Show that $x_n \leq \frac{1}{\sqrt[3]{n-1}}$ for each $n\geq2$.

$(ii)$ $[\mathbf1]$ Prove that every integer $n\geq2$ satisfies $\frac{1}{2^{2/3}} + \frac{1}{3^{2/3}} + \dots + \frac{1}{n^{2/3}} \leq \int_1^nx^{-2/3}dx$, and hence show that $x_{n+1} > \frac{1}{3\sqrt[3]{n-1}}$ for each integer $n\geq2$.

$(iii)$ $[\mathbf1]$ Prove that every $n\geq2$ also satisfies $\frac{1}{\sqrt[3]{1}} + \frac{1}{\sqrt[3]{2}} + \dots + \frac{1}{\sqrt[3]{n}} \geq \int_1^{n+1}\frac{1}{\sqrt[3]{x}}dx$, and hence deduce that $x_1 + x_2 + \dots + x_n > \frac{(n-1)^{2/3}}{2}$ for each $n\geq2$.

$(iv)$ $[\mathbf1]$ By considering a similar integral in $(iii)$, show that for all sufficiently large positive integers $n$ the inequality $\frac{1 - \frac{1}{2019}}{2} < \frac{x_1 + x_2 + \dots + x_n}{n^{2/3}} < \frac{3 + \frac{1}{2019}}{2}$ holds.

$(v)$ $[\mathbf2]$ It is given ($\mathbf{Do}$ $\mathbf{NOT}$ $\mathbf{prove}$) that the limit $\lim\limits_{n \to \infty} \frac{x_1 + x_2 + \dots + x_n}{n^{2/3}}$ exists, and is equal to a positive real number $C$.

Show that $C = \frac{\sqrt[3]{9}}{2}$.

Attempt so far

I was able to solve parts (i), (ii) and (iii) and am able to solve the lower bound in (iv) but cannot find a good integral for the upper bound. (My problem is trying to make n+1 appear in the integrand, while also only using n Riemann sums for the bound). I also see issues with applying squeeze law in (v).

Request For Hint

What is a good hint for finding the correct integral in (iv), and how do I extend this to part (v) where I'm hoping squeeze law applies but can't see a trivial path forward if the two integral bounds have multipliers of $\frac{1}{2}$ and $\frac{3}{2}$ respectively (thereby squeezing to a range rather than a point).

Working for part (iii)

And an asymptotic hack to get out part (iv)

Answer 1

This is my first time answering question on this website.If my editing form is wrong,please forgive me.Thanks. $$ x_{n+1}=\frac{1}{{x_1}^2+{x_2}^2+...+{x_{n-1}}^2+{x_n}^2} \\ so\,\,x_n=\frac{1}{{x_1}^2+{x_2}^2+...+{x_{n-1}}^2}\,\,and\,\,\frac{1}{x_n}={x_1}^2+{x_2}^2+...+{x_{n-1}}^2 \\ so\,\,x_{n+1}=\frac{1}{\frac{1}{x_n}+{x_n}^2}=\frac{x_n}{1+{x_n}^3}\,\,We\,\,got\,\,the\,\,recurrence\,\,relation. \\ \,\,x_1=1 ,according\,\,to\,\,recurrence\,\,relation, we\,\,know\,\,x_{n+1}-x_n<0, x_n\rightarrow 0\left( n\rightarrow \infty \right) \\ \frac{1}{{x_{n+1}}^3}=\frac{1+3{x_n}^3+3{x_n}^6+{x_n}^9}{{x_n}^3}=\frac{1}{{x_n}^3}+3+O\left( {x_n}^3 \right) \\ so\,\, \frac{1}{{x_{n+1}}^3}-\frac{1}{{x_n}^3}=3+O\left( {x_n}^3 \right) \\ so\,\,\frac{1}{{x_{n+1}}^3}-\frac{1}{{x_1}^3}=3n+\sum_{n=1}^n{O\left( {x_n}^3 \right)} \\ \frac{1}{{x_{n+1}}^3}=3n+\sum_{n=1}^n{O\left( {x_n}^3 \right)}+\frac{1}{{x_1}^3} \\ x_{n+1}\sim \left( \frac{1}{3n} \right) ^{\frac{1}{3}} \\ \lim_{n\rightarrow \infty} \frac{x_1+x_2+...x_{n+1}}{n^{\frac{2}{3}}}\overset{OStolz\,\,therom}{=}\lim_{n\rightarrow \infty} \frac{x_{n+1}}{\left( n+1 \right) ^{\frac{2}{3}}-n^{\frac{2}{3}}}=\lim_{n\rightarrow \infty} \frac{\left( \frac{1}{3n} \right) ^{\frac{1}{3}}}{n^{\frac{2}{3}}\left( \left( 1+\frac{1}{n} \right) ^{\frac{2}{3}}-1 \right)}=\lim_{n\rightarrow \infty} \frac{\left( \frac{1}{3n} \right) ^{\frac{1}{3}}}{n^{\frac{2}{3}}\frac{2}{3}\frac{1}{n}}=\frac{\sqrt[3]{9}}{2} $$