<s>[INST] Prove the series generated by the sequence diverges Consider the positive sequence $\{x_n\}_{n\ge 1}$ given by
$$\begin{cases}
x_1&=1,\\
x_{n+1}^2+x_{n+1}x_n^2-x_n^2&=0,\ \forall n\ge 1.
\end{cases}$$
More specifically, the second equation together with condition that the sequence is positive-valued mean that $$ x_{n+1}=\dfrac{-x_n^2+\sqrt{x_n^4+4x_n^2}}{2}>0.$$
Prove that $\sum_{n=1}^\infty x_n=+\infty$.
With some ideas from Kaind, it can be seen that
\begin{align*}
    x_{n+1}< x_n\Longleftrightarrow -x_n^2+\sqrt{x_n^4+4x_n^2}< 2x_n\Longleftrightarrow x_n^4+4x_n^2 < (x_n^2+2x_n)^2=x_n^4+4x_n^3+4x_n^2.
\end{align*}
Since the latter inequality is true, we have $x_{n+1}< x_n$ for all $n,$ which means $x_n$ is decreasing. Since the sequence is bounded from below by $0,$ its limit exists. Assume the limit is $L,$ we have $L^2+L^3-L^2=0,$ or equivalently, $L=0.$
Thus $x_n\downarrow 0$. It remains to show that $\sum_{n=1}^\infty x_n=+\infty$ and I am still stucking here.
 [/INST] This is the answer from one of my friends.
We will prove that $x_n\ge 1/n$ by induction. Firstly, notice that
\begin{align*}
    x_{n+1}=\dfrac{-x_n^2+\sqrt{x_n^4+4x_n^2}}{2}=\dfrac{4x_n^2}{2(x_n^2+\sqrt{x_n^4+4x_n^2})}=\dfrac{2}{1+\sqrt{1+\dfrac{4}{x_n^2}}}.
\end{align*}
Consider the function $f(x)=\dfrac{2}{1+\sqrt{1+\dfrac{4}{x^2}}}$, we have $x_{n+1}=f(x_n).$ Since $f$ is increasing, supposing $x_n\ge 1/n$, we have
\begin{align*}
    x_{n+1}=f(x_n)\ge f(1/n)=\dfrac{2}{1+\sqrt{1+4n^2}}.
\end{align*}
It remains to show that $\dfrac{2}{1+\sqrt{1+4n^2}}\ge \dfrac{1}{n+1}$ which is equivalent to
\begin{align*}
    2n+2\ge 1+\sqrt{1+4n^2}\Longleftrightarrow 2n+1\ge \sqrt{1+4n^2}.
\end{align*}
Squaring both side gives us a always-true expression $4n\ge 0$, and thus $x_{n+1}\ge \dfrac{1}{n+1}.$ Together with $x_1\ge 1$, by induction we have $x_n\ge 1/n$ for all $n.$ Thus $\sum_{n=1}^\infty x_n=\infty.$
</s>