Let $\langle x_n\rangle$ be a recursive relation. Find $\lim_{n\to\infty}\frac {x_n}{n^2}.$ Let $\langle x_n\rangle$ be a recursive relation given by $$x_{n+1}=x_n+a+\sqrt {b^2+4ax_n}, n\geq0, x_0 =0$$ and $a$ and $b$ are fixed positive integers. Find  $$\lim_{n\to\infty}\frac {x_n}{n^2}.$$
 A: Clearly, $\lim_{n\to \infty} x_n = \infty$. We have
\begin{align}
\sqrt{x_{n+1}} - \sqrt{x_n} &= \sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} - \sqrt{x_n}\\[6pt]
&= \frac{a + \sqrt{b^2 + 4ax_n}}{\sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} + \sqrt{x_n}}\\[6pt]
&\to \sqrt{a} \quad \mathrm{as}\quad n \to \infty.
\end{align}
By the Stolz-Cesaro theorem, we have
$$\lim_{n\to \infty} \frac{\sqrt{x_n}}{n} = \sqrt{a}$$
and thus
$$\lim_{n\to \infty} \frac{x_n}{n^2} = a.$$
We are done.
A: I attempted it like this. ( Although it isn't a 'proper' method ) :-
$$x_0=0$$$$x_1=0+a+\sqrt {b^2+4a(0)}=a+b$$$$x_2=(a+b)+a+\sqrt {b^2+4a(a+b)}=4a+2b$$$$x_3=(4a+2b)+a+\sqrt {b^2+4a(4a+2b)}=9a+3b$$$$\cdots$$$$x_n=an^2+bn$$$$\therefore \lim_{n\to\infty}\frac{x_n}{n^2}=\lim_{n\to\infty}a+\frac bn=a.$$
A: It is practical to set $x_n=a z_n$ and $r=\frac{b}{a}$ to get rid of a parameter:
$$ z_{n+1} = z_{n} + 1 + \sqrt{4z_n+r^2},\qquad z_0=0 $$
Obviously $z_n\geq n$, since $z_{n+1}\geq z_{n}+1$, but also
$$ z_{n+1} \geq (\sqrt{z_n}+1)^2, $$
hence by setting $z_{n}=w_n^2$ we get $w_{n+1}\geq w_n+1$ and $z_n\geq n^2$. On the other hand
$$ z_{n+1}-(\sqrt{z_n}+1)^2 = \sqrt{4z_n+r^2}-\sqrt{4z_n} = \frac{r^2}{\sqrt{4z_n}+\sqrt{4z_n+r^2}}\leq \frac{r^2}{4n} $$
$$ w_{n+1}^2 - (w_n+1)^2 \leq \frac{r^2}{4n} $$
$$ w_{n+1}-(w_n+1) \leq \frac{r^2}{4n(w_{n+1}+w_n+1)} \leq \frac{r^2}{8n(n+1)} $$
hence $w_n=n+O(1)$, $z_n=n^2+O(n)$ and $x_n = \color{red}{a}n^2+O(n)$.
