Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$

Question

$a\gt0$, 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!

Answer 1

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 $C_1$.

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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.

Answer 2

Here is an idea:

Let $\epsilon_n=a+1-x_n$ so that $\epsilon_0=a+1$ and $$(a+1-\epsilon_{n+1})^2=a+1-\epsilon_n+a(a+1)$$ or$$(a+1)^2-2\epsilon_{n+1}(a+1)+(\epsilon_{n+1})^2=(a+1)^2-\epsilon _n$$ whence $$2\epsilon_{n+1}(a+1)-(\epsilon_{n+1})^2=\epsilon_n$$

Answer 3

First it is easy to show that $x_n$ increases and $x_n\to a+1$.

As hinted by @user10676, let's consider the following function around $x=a+1$ $$ f(x)=\sqrt{x+a(a+1)}=f(a+1)+f'(a+1)(x-a-1)+etc\\\sim a+1+\frac{1}{2(a+1)}(x-a-1). $$ Since $x_n\to a+1$, $$ x_{n+1}-(a+1)\sim\frac{1}{2(a+1)}(x_n-a-1) $$ So we deduce that $(a+1-x_n)\sim\frac{c}{(2(a+1))^n}$ for some constant to be determined (I am not able to determine it for the moment). I suppose there is a typo in the text.

If there is no typo, then the result is either $0$ or $\infty$ depending on $a<1,a>1$. When $a=1$, we need to evaluate

$$ \frac{4}{2}\frac{4}{2+\sqrt{2}}\frac{4}{2+\sqrt{2+\sqrt{2}}}\cdots\frac{4}{2+\sqrt{2+\sqrt{2+\cdots}}}\cdots, $$ see Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$

Answer 4

I guess the answer in zero. It's enough to check that $$\lim_{n\to\infty}x_n=a+1$$ This makes the second part of your limit zero.
In order to compute that limit, let's assume it exists and it's $L$.
We have $x_{n+1}^2=x_n+a^2+a$
then $x_{n+1}^2-x_n-a^2-a=0$
taking limits we have $L^2-L-a^2-a=0$
solving in $L$ there is only one valid solution $L=a+1$

Does it make sense?.

Comment on the above This is not enough as limit becomes in the form $\infty * 0$, I have to make a refinement on the proof. Empirically I have checked the result is zero.