Question

Let $a_n$ be a sequence of real numbers with the initial condition $a_0 = 0$ and the recurrence relation $a_{n+1} = a_n + \sqrt{1 + a_n^2}$.

Find the value of the limit: $\lim_{n \to \infty} (\frac{a_n}{2^n})$

Answer 1

$\frac{2}{\pi}$

Answer 2

The problem asks us to find the limit of the sequence $\frac{a_n}{2^n}$ as $n \to \infty$, given the initial condition $a_0 = 0$ and the recurrence relation $a_{n+1} = a_n + \sqrt{1 + a_n^2}$.

1. Transform the recurrence relation using a trigonometric substitution:

Let $a_n = \tan \theta_n$. Since $a_0 = 0$, $a_1 = 1$, and all subsequent terms will be positive, we can assume $\theta_n \in [0, \pi/2)$. Substitute $a_n = \tan \theta_n$ into the recurrence relation: $a_{n+1} = \tan \theta_n + \sqrt{1 + \tan^2 \theta_n}$ Since $1 + \tan^2 \theta_n = \sec^2 \theta_n$ and $\theta_n \in [0, \pi/2)$, $\sec \theta_n > 0$. So, $\sqrt{1 + \tan^2 \theta_n} = \sec \theta_n$. $a_{n+1} = \tan \theta_n + \sec \theta_n$ $a_{n+1} = \frac{\sin \theta_n}{\cos \theta_n} + \frac{1}{\cos \theta_n} = \frac{1 + \sin \theta_n}{\cos \theta_n}$

Now, we use trigonometric identities involving half-angles: $1 + \sin \theta_n = 1 + \cos(\frac{\pi}{2} - \theta_n) = 2 \cos^2(\frac{\pi}{4} - \frac{\theta_n}{2})$ $\cos \theta_n = \sin(\frac{\pi}{2} - \theta_n) = 2 \sin(\frac{\pi}{4} - \frac{\theta_n}{2}) \cos(\frac{\pi}{4} - \frac{\theta_n}{2})$ Substitute these into the expression for $a_{n+1}$: $a_{n+1} = \frac{2 \cos^2(\frac{\pi}{4} - \frac{\theta_n}{2})}{2 \sin(\frac{\pi}{4} - \frac{\theta_n}{2}) \cos(\frac{\pi}{4} - \frac{\theta_n}{2})} = \cot(\frac{\pi}{4} - \frac{\theta_n}{2})$ Using the identity $\cot x = \tan(\frac{\pi}{2} - x)$: $a_{n+1} = \tan\left(\frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{\theta_n}{2}\right)\right) = \tan\left(\frac{\pi}{4} + \frac{\theta_n}{2}\right)$

Since $a_{n+1} = \tan \theta_{n+1}$, we have the recurrence relation for $\theta_n$: $\theta_{n+1} = \frac{\pi}{4} + \frac{\theta_n}{2}$

2. Solve the linear recurrence relation for $\theta_n$:

This is a linear recurrence relation of the form $x_{n+1} = A + B x_n$. The fixed point $\theta^*$ is found by setting $\theta^* = \frac{\pi}{4} + \frac{\theta^*}{2}$, which gives $\frac{\theta^*}{2} = \frac{\pi}{4}$, so $\theta^* = \frac{\pi}{2}$. The general solution is $\theta_n = \theta^* + (\theta_0 - \theta^*) B^n$. We have $a_0 = 0$, so $\tan \theta_0 = 0$. We choose $\theta_0 = 0$. Substitute the values: $\theta_n = \frac{\pi}{2} + (0 - \frac{\pi}{2}) \left(\frac{1}{2}\right)^n = \frac{\pi}{2} - \frac{\pi}{2^{n+1}}$

3. Express $a_n$ in terms of $n$:

Now substitute $\theta_n$ back into $a_n = \tan \theta_n$: $a_n = \tan\left(\frac{\pi}{2} - \frac{\pi}{2^{n+1}}\right)$ Using the identity $\tan(\frac{\pi}{2} - x) = \cot x$: $a_n = \cot\left(\frac{\pi}{2^{n+1}}\right)$

4. Evaluate the limit:

We need to find $\lim_{n \to \infty} \left(\frac{a_n}{2^n}\right)$. Substitute the expression for $a_n$: $\lim_{n \to \infty} \left(\frac{\cot\left(\frac{\pi}{2^{n+1}}\right)}{2^n}\right)$ Let $x_n = \frac{\pi}{2^{n+1}}$. As $n \to \infty$, $x_n \to 0$. The term $2^n$ can be expressed in terms of $x_n$: $2^{n+1} = \frac{\pi}{x_n} \implies 2^n = \frac{1}{2} \cdot \frac{\pi}{x_n} = \frac{\pi}{2x_n}$. Substitute this into the limit expression: $\lim_{n \to \infty} \left(\frac{\cot x_n}{\frac{\pi}{2x_n}}\right) = \lim_{n \to \infty} \left(\frac{2x_n \cot x_n}{\pi}\right)$ This can be rewritten as: $\frac{2}{\pi} \lim_{n \to \infty} \left(x_n \cot x_n\right) = \frac{2}{\pi} \lim_{x_n \to 0} \left(\frac{x_n}{\tan x_n}\right)$ We know the standard limit $\lim_{x \to 0} \frac{\tan x}{x} = 1$. Therefore, $\lim_{x \to 0} \frac{x}{\tan x} = 1$. So, the limit is: $\frac{2}{\pi} \cdot 1 = \frac{2}{\pi}$