Let $a_i (i \in \mathbb{N}_{0})$ be a sequence of real numbers such that $a_0 = 0$ and $$a_n = a_{n – 1} + \sqrt{a_{n – 1}^2 + 1} \text{ } \forall n \geq 1$$ Evaluate the limit $$\lim_{n \to \infty} \frac{a_n}{2 ^ {n – 1}}$$
Hello, I am trying to solve this problem. I honestly have no idea how to approach this, but I think the answer will be $1$ because as $n \to \infty$, $a_n$ gets bigger and $a_n \approx 2a_{n-1}$ and since $a_1 = 1 = 2^{0}$, the limit will approach $1$. (Of course, this is just a guess).
The intended solution is too much magic.
Substitute $a_n = \cot \theta$.
Now $$a_{n + 1} = a_n + \sqrt{a_{n}^2 + 1} = \cot \theta + \csc \theta$$
$$=\frac{\cos \theta + 1}{\sin \theta}$$
$$=\frac{2 \cos^2{\frac{\theta}{2}}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$$
$$=\cot \frac{\theta}{2}$$
And now, we can solve the limit and answer is $\frac{4}{\pi}$.
But, is there a more normal method to solving this (other than just thinking out of nowhere that substituting $a_n = \cot \theta$ is helpful)?
Please note that my question is about finding an alternative solution that's much more "thinkable". So, this is not a duplicate.
Thanks
Best Answer
This is a general method that you may be able to use.
Suppose given a function
$$f(x) := x + \sqrt{x^2+1}. $$
Given a number $\,a_0\ge 0,\,$ define the sequence
$$a_n := f(a_{n-1}). $$
Note the series expansion $$ f(x) = 2x + \frac1{2x} - \frac1{8x^3} + \frac1{16x^5} + \cdots. $$
Iterate to get
$$ f^2(x) := f(f(x)) = 4x\left( 1 +\frac5{16x^2} -\frac{21}{256x^4} + \cdots\right). $$ In general,
$$ f^n(x) := f(f^{n-1}(x)) = 2^nx\left( 1 + \frac{c_n}{x^2} + \frac{d_n}{x^4} + \cdots\right). $$
With a bit of work, take the limit to get
$$ g(x) := \lim_{n\to\infty} \frac{f^{n}(x)}{2^n} = x\left(1 +\frac{1/3}{x^2} - \frac{4/45}{x^4} + \frac{44/945}{x^6} + \cdots\right). $$
In this particular case,
$$ \frac1{g(x)} = \frac1{x} - \frac1{3x^3} + \frac1{5x^5} - \frac1{7x^7} + \cdots = \tan^{-1}(1/x).$$
So define $\,b_n := a_{n+1}\,$ where $\,b_0 = 1.\,$ Then
$$ \lim_{n \to \infty} \frac{a_n}{2^{n-1}} = \lim_{n \to \infty} \frac{a_{n+1}}{2^n} = \lim_{n \to \infty} \frac{b_n}{2^n} = 1/\tan^{-1}(1/b_0) = 4/\pi.$$
The purpose of this section is to derive the equation
$$ \frac1{g(x)} = \tan^{-1}(1/x) $$
which appears near the end of the previous section.
The general idea is to study discrete dynamical systems and their limiting behavior. Thus, given a subset $\,X\,$ of real numbers, a self-map $\,\phi:X\to X,\,$ and a point $\,a_0\,$ in $\,X\,$ define the sequence $\,(a_0,a_1,a_2, \dots)\,$ by recursion with $\,a_{n+1} = \phi(a_n)\,$ for $n>0.\,$
Another way to think of this situation is from the standpoint of iterated functions. That is, given a self-map $\,\phi:X\to X,\,$ define a sequence of functions by recursion as $\,\phi^0: x \mapsto x,\,$ and $\,\phi^{n+1}: x \mapsto \phi(\phi^n(x)).\,$ Thus, the sequence $\,a_n\,$ is also defined by $\,a_n = \phi^n(a_0).\,$ We are interested in fixed points of the self-map and convergence to fixed points.
In our particular case, $\,f(x) = x+\sqrt{x^2+1}\,$ and the sequence $\,a_n\,$ is unbounded. Thus, it is more convenient to define
$$ \phi(x) := \frac1{f(1/x)} = \frac{x}{1+\sqrt{1+x^2}} = \frac{x}2 - \frac{x^3}8 + \frac{x^5}{16} - \frac{5x^7}{128} + \cdots $$
This implies that iterates of $\,\phi\,$ converge to the fixed point $0$. The convergence to this limit is linear. More precisely, define
$$ A_t(x) := x\,t\left(1 - x^2\frac{1-t^2}3\left(1 - x^2\frac{3-2t^2}5 + x^4\frac{45-53t^2+17t^4}{105} + \cdots\right)\right)$$
Verify that this function satisfies the identities
$$ A_u(A_t(x)) = A_{ut}(x),\qquad A_{1/2}(x) = \phi(x). $$
The relation of this function to the sequence $\,b_n\,$ is $\, b_n = 1/A_{2^{-n}}(b_0).\,$
We seek another function, $\,T(x),\,$ which "conjugates" multiplication. More precisely,
$$ A_t(x) = T(t\,T^{-1}(x)), \quad T^{-1}(x) = x - \frac{x^3}3 + \frac{x^5}5 + \cdots $$
Thus, $\,A_t(x)\,$ is completely determined by the function $\,T(x).\,$
The special case of $\,t=2,\,$ implies $\, T^{-1}(\phi^{-1}(x)) = 2\,T^{-1}(x).\,$ Differentiate both sides to get
$$ {T^{-1}}'(\phi^{-1}(x))\cdot \frac{2\,(1+x^2)}{(1-x^2)^2} = 2\,{T^{-1}}'(x).$$
Define $\,U(x) := {T^{-1}}'(x)\,$ and simplify to get
$$ U(\phi^{-1}(x)) \frac{(1+x^2)^2}{(1-x^2)^2} = U(x)\,(1+x^2). $$
Substitute $\,\phi(x)\,$ for $\,x\,$ to get
$$ U(x)\frac{(1+\phi(x)^2)^2}{(1-\phi(x)^2)^2} = U(\phi(x))\,(1+\phi(x)^2). $$
Use the definition of $\,\phi(x)\,$ in the left side of the equation to get
$$ U(x)\,(1+x^2) = U(\phi(x))\,(1+\phi(x))^2. $$
Because this holds for all $\,x,\,$ both sides are equal to the constant $\,U(0)=1.\,$ This implies
$$ U(x) = \frac1{1+x^2}, \qquad T^{-1}(x) = \tan^{-1}(x). $$