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My professor provided a solution to this problem, but it seemed very unmotivated. I am looking for a general way to approach recursively defined sequences and show their convergence. Here is the specific question. Let $x_0 = 1$. Let $x_{n+1} = \frac{x_n +2}{x_n +1}$ for $n \in \mathbb{N} \cup {0}$. Show that $x_n$ converges to $\sqrt{2}$. Edit: I should add that computing the limit as $\sqrt{2}$ wouldn't be hard if I could show that the limit in fact exists, but the only way I could think of to do that would be to show that this was a bounded increasing sequence, but the sequence isn't monotonically increasing as far as I can show.

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Consider an iteration $x_{n+1} = f(x_n)$ where $f$ is some function. If $\lim_{n \to \infty} x_n = L$ and $f$ is continuous at $L$, then $L$ must be a fixed point of $f$, i.e. $f(L) = L$. Moreover, if $f$ is differentiable at $L$, $L$ is an attracting fixed point of $f$ if $|f'(L)|< 1$, while if $|f'(L)| > 1|$ it is a repelling fixed point. An attracting fixed point will be the limit of the sequence if the terms ever get close enough, while a repelling fixed point can't be the limit unless some $x_n$ happens to be exactly $L$.

One way to show that the fixed point $L$ is the limit of the sequence is to find an interval $[a,b]$ containing $L$ and such that $|f'(x)| < 1$ for all $x$ in $L$, and show that some $x_n$ is in the interval.

In your case, $f(x) = \frac{x+2}{x+1}$ has two fixed points, one attracting and one repelling.

EDIT: In this particular case, another approach is possible, since this function is a fractional linear transformation $f(x) = \frac{ax + b}{cx+d}$. For these, it turns out that $x_n = \frac{a_n x_0 + b_n}{c_n x_0 + d_n}$ where the matrix $$\pmatrix{a_n & b_n\cr c_n & d_n} = \pmatrix{a & b\cr c & d\cr}^n$$ We can then use a little linear algebra to get closed-form formulas for $x_n$. In your case, the matrix has eigenvalues $\lambda_1 = 1+\sqrt{2}$ and $\lambda_2 = 1 - \sqrt{2}$ corresponding to eigenvectors $\pmatrix{\pm \sqrt{2} \cr 1\cr}$, so that $$ \pmatrix{1 & 2\cr 1 & 1\cr}^n = \pmatrix{\sqrt{2} & -\sqrt{2}\cr 1 & 1} \pmatrix{\lambda_1^n & 0\cr 0 & \lambda_2^n} \pmatrix{ \sqrt{2} & -\sqrt{2} \cr 1 & 1\cr}^{-1} = \pmatrix{(\lambda_1^n+\lambda_2^n)/2 & \sqrt{2}(\lambda_1^n-\lambda_2^n)/2\cr \sqrt{2}(\lambda_1^n-\lambda_2^n)/4 & (\lambda_1^n + \lambda_2^n)/2} $$ Thus with $x_0 = 1$, $$ x_n = \frac{(\lambda_1^n+\lambda_2^n)/2 + \sqrt{2}(\lambda_1^n-\lambda_2^n)/2}{ \sqrt{2}(\lambda_1^n-\lambda_2^n)/4 + (\lambda_1^n + \lambda_2^n)/2} = \frac{(1+\sqrt{2})/2 + (1-\sqrt{2}) (\lambda_2/\lambda_1)^n/2}{(\sqrt{2}/4 + 1/2) + (-\sqrt{2}/4 + 1/2) (\lambda_2/\lambda_1)^n}$$ Since $|\lambda_2| < |\lambda_1|$, $(\lambda_2/\lambda_1)^n \to 0$ as $n \to \infty$, so that $$x_n \to \frac{(1+\sqrt{2})/2}{\sqrt{2}/4+1/2} = \sqrt{2}$$

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  • $\begingroup$ Any way to do this without any machinery of continuity or derivation? $\endgroup$ – Nujra Nov 16 '17 at 6:46
  • $\begingroup$ @Nujra Professional mathematicians use professional tools. If you don't want to learn them, don't go looking for "a general way to approach recursively defined sequences". $\endgroup$ – Professor Vector Nov 16 '17 at 6:53
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Based on the theoretical knowledge given by R.Israel for sequences $(x_n)_n$ defined by $x_{n+1}=f(x_n)$ you can also draw the graph of $y=x$ and $y=f(x)$ and notice that for $x_0=1$ the sequence $x_n$ converges to $\ell=\sqrt{2}$ in spiral.

Now, instead of going for calculating $f'$ you can try to justify these visual observations.

First note that $x\ge 1\implies f(x)\ge 1$, so since $x_0=1$ then $x_n\ge 1$ for any $n$.

Since it is a spiral convergence, we are interested in the behaviour of $x_{n+2}-x_n$.

So let's study

  • $f(f(x))-x=\dfrac{4-2x^2}{2x+3}=\underbrace{\dfrac{-2}{2x+3}}_{<0}(x^2-\ell^2)$

  • $f(x)-\ell=\underbrace{\dfrac{\sqrt{2}-1}{x+1}}_{>0}(x-\ell)$

Whose signs depends only whether $x>\ell$ or not considering we already have $x\ge 1$.

Let's define : $\begin{cases} u_n=x_{2n} & u_0=x_0=1<\ell\\v_n=x_{2n+1} & v_0=x_1=\frac 32>\ell\end{cases}\implies\begin{cases}(u_n)_n\nearrow & u_n<\ell\\(v_n)_n\searrow & v_n>\ell\end{cases}$

So both sequences are convergent and the equation $f(f(x))-x=0$ immediately confirms that this limit is common and actually $\ell$.

Yet, as you can notice, this is no easiest calculations than the $f'$ based approach, just another way to present it.

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