<s>[INST] Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$ $a\gt0$, and  $x_0=0$, $x_{n+1}=\sqrt{x_n+a(a+1)}, n=0,1,2,\dotsc$.
compute
$$\lim_{n\to\infty}\big(a+1\big)^{2n}\big(a+1-x_n\big)$$
The problem is difficult, I have no idea.
Thank you!
 [/INST] From the recursion given, we get
$$
x_{n+1}^2=x_n+a(a+1)\tag{1}
$$
Subtracting $(1)$ from $(a+1)^2$ gives
$$
(a+1)^2-x_{n+1}^2=(a+1)-x_n\tag{2}
$$
which is equivalent to
$$
\frac{(a+1)-x_{n+1}}{(a+1)-x_n}
=\frac1{(a+1)+x_{n+1}}\tag{3}
$$
which inductively implies
$$
(a+1)-x_n\le\frac{(a+1)-x_0}{(a+1)^n}\tag{4}
$$
Multiplying $(3)$ by $2(a+1)$ gives
$$
\begin{align}
2(a+1)\frac{(a+1)-x_{n+1}}{(a+1)-x_n}
&=\frac{2(a+1)}{(a+1)+x_{n+1}}\\
&=1+\frac{(a+1)-x_{n+1}}{(a+1)+x_{n+1}}\tag{5}
\end{align}
$$
which inductively implies
$$
2^n(a+1)^n((a+1)-x_n)=((a+1)-x_0)\prod_{k=1}^n\left(1+\frac{(a+1)-x_k}{(a+1)+x_k}\right)\tag{6}
$$
Inequality $(4)$ guarantees that the product in $(6)$ converges. Thus, there is a $0\lt C_a\lt\infty$ so that
$$
\lim_{n\to\infty}2^n(a+1)^n((a+1)-x_n)=C_a\tag{7}
$$
Therefore, if $a\gt1$, then the given limit is $\infty$, and if $a\lt1$, then the given limit is $0$. If $a=1$, then the given limit is $C_1$.

Computation of $C_1$
When $a=1$, recursion $(1)$ becomes
$$
x_{n+1}^2=x_n+2\tag{8}
$$
which is satified by
$$
x_n=2\cos(2^{-n}t_0)\tag{9}
$$
with $t_0=\frac\pi2$. Therefore,
$$
\begin{align}
C_1
&=\lim_{n\to\infty}4^n(2-2\cos(2^{-n}t_0))\\
&=t_0^2\\
&=\frac{\pi^2}{4}\tag{10}
\end{align}
$$
As has been mentioned, this is related to this answer.
</s>