If $x_{n+1}=x_n+\sqrt{x_n^2-1}$ for $n\in\mathbb W$ and $x_0=2$, then evaluate $\lim_{n\to\infty}2^{-n}x_n$.

Question

If $x_{n+1}=x_n+\sqrt{x_n^2-1}$ for $n\in\mathbb W(\mathbb W$ denotes the set of whole numbers$)$ and $x_0=2$, then evaluate $\displaystyle\lim_{n\to\infty}2^{-n}x_n$.

This is a self-answered post. Any alternative methods/comments are highly appreciated.

Motivation/Context: I thought of this question by modifying one of the questions that appeared on my sequences and series test, which is:

If $x_{n+1}=\dfrac{x_n}{1+\sqrt{x_n^2+1}}$ for $n\in\mathbb W$ and $x_0=\sqrt3$, then compute $\displaystyle\lim_{n\to\infty}2^nx_n$.

$($which can be solved by recognizing the identity $\tan\frac\theta2=\frac{\tan\theta}{1+\sec\theta}$ for $\theta\ge0)$

This is my first self-answered post. If it is missing any details or needs any improvements, kindly comment. — Poonguzhali Annadurai, Mar 20 '25 at 18:17
Welcome to Math SE. To me, at least, your post looks fine. FYI, using an Approach0 search, just in somewhat more than the first page of results, there are many posts and on AoPS regarding the similar recurrence relation of $x_{n+1}=x_n+\sqrt{x_n^2+1}$. In particular, on this site, there's Find limit n tends to infinity, The convergence of a recursive sequence, ... — John Omielan, Mar 20 '25 at 19:16
(cont.) Limit $\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$ for $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$, How to evaluate $\lim_{ n\to \infty }\frac{a_n}{2^{n-1}}$, if $a_0=0$ and $a_{n+1}=a_n+\sqrt{a_n^2+1}$?, Find the limit $\lim_{n\to \infty} k_n/2^n$ for $k_1=0$ and $k_{n+1}=k_n+\sqrt{1+k_n^2}$, etc. The AoPS threads include Recursive limit, Limit Prob Please solve, ... — John Omielan, Mar 20 '25 at 19:19
why do you have $n\in\mathbb W ; ; ; ? ; ; $ I would expect $n\in\mathbb N$ — Will Jagy, Mar 20 '25 at 19:34
@IntegralKaadhalan, what do you mean when you write $\Bbb W$ ? — Angelo, Mar 21 '25 at 02:15
@Angelo it denotes the set of whole numbers. I don’t know about other countries, but in Indian schools, this notation is common. — Poonguzhali Annadurai, Mar 21 '25 at 04:16
@IntegralKaadhalan, $\Bbb Z$ denotes the set of integers, for example, $0$, $1$, $-1$, $2$, $-2$, $3$, $-3$, etc. Moreover, we use $\Bbb N$ for denoting the set of positive integers, for example, $1$, $2$, $3$, etc. And we use $\Bbb N_0$ for denoting the set of non-negative integers, for example, $0$, $1$, $2$, $3$, etc. Perhaps, you denote by using $\Bbb W$ what we denote by using $\Bbb N_0$, do not you? — Angelo, Mar 21 '25 at 12:19
@Angelo yes, you're right. BTW, I've never seen $\mathbb N_0$ before. — Poonguzhali Annadurai, Mar 21 '25 at 14:10
@IntegralKaadhalan, I can assure you that actually I have never seen $\Bbb W$ before today. Please, look at this. — Angelo, Mar 21 '25 at 14:48

score 2 · Accepted Answer · answered Mar 20 '25 at 18:12

The given recurrence relation is reminiscent of the identity

$$\coth\frac{x}2=\coth x+\operatorname{csch}x=\coth x+\operatorname{sgn}(x)\sqrt{\coth^2x-1}$$

Hence, substituting $x_n=\coth\theta$, we get $x_{n+1}=\coth\frac\theta2$. So,

$$x_n=\coth\left(\frac{\coth^{-1}x_0}{2^n}\right)$$ $$\implies\lim_{n\to\infty}2^{-n}x_n=\frac1{\coth^{-1}x_0}$$

Hence, if $x_0=2$, then the value of the given limit is $\frac2{\ln3}$.

If $x_{n+1}=x_n+\sqrt{x_n^2-1}$ for $n\in\mathbb W$ and $x_0=2$, then evaluate $\lim_{n\to\infty}2^{-n}x_n$.

1 Answers1