I'm kinda new to recursive sequences and I'm struggling with an excercise. I apologize in advance for the long question and my lack of knowledge on how to approach such problems.
Let $x_n$ be defined such that $x_1 = 1$, $x_{n+1} = \frac{x_n^2+1}5$
And there there are 4 things I have to prove/show,
(1) Show that $x_n$ $\leq$ $1$ for all $n$.
I'm not sure if I can prove this directly or by induction? What would be best for this? I don't know how to approach this when it's defined recursively.
(2) Show that $\lvert x_{n+1} – x_n \rvert$ $\leq (\frac{2}{5})^{n-1}$
I'm not sure but could I use $\lvert\frac{x_{n+1}^2+1}{5}$ $-$ $\frac{x_n^2+1}{5}\rvert$$\leq (\frac{2}{5})^{n-1}$ and then try to solve this. How do i deal with the $x_n$?
(3) Show that $x_n$ is a Cauchy-sequence.
I know that a Cauchy sequence is defined as : For every positive real number $\varepsilon$, there is a positive integer $N$ such that for all natural numbers $m, n \gt N$: $\lvert x_m – x_n \rvert \lt \varepsilon$. Can I apply this definition directly in this sequence and the go on from there?
(4) Calculate the limit of $x_n$.
This should be straightforward if it weren't a recursively defined sequence.
If I however, calculate the first few elements of the sequence I get:
$x_1 = 1$, $x_2 = \frac{2}{5}$, $x_3 = \frac{29}{125}$, $x_4 = \frac{841}{15625}$ $\ldots$
This approaches $0$ (I'm guessing) since the second term is smaller than $1$. And since a square of a number less than $1$ is always smaller than the original number, $\lim \limits_{n \to \infty}$ $= 0$.
How can I prove this correctly without just trying different numbers?
Thank you in advance!
Best Answer
Since $x_1 \leq 1$, we just need to prove that, if $x_n \leq 1$, then $x_{n+1}\leq 1$. This is straightforward because we have that $x_{n+1} = \dfrac{x_n^2+1}{5}\leq \frac 25 \leq 1$. Also, considering that $x_n >0$, we have established that $0 < x_n \leq 1, \forall_n$.
$$|x_{n+1} - x_n| = \left| \frac{x_n^2+1}{5} - \frac{x_{n-1}^2+1}{5} \right|=\left| \frac{(x_n-x_{n-1})(x_n+x_{n-1})}{5}\right| \leq \frac 25 |x_n -x_{n-1}|$$ Using this inequality recursively, one can realize that $$|x_{n+1}-x_n|\leq (2/5)^{n-1}|x_2-x_1| \leq (2/5)^{n-1}.$$
The sequence is Cauchy if, for every $\varepsilon >0$, there is some order $p$ such that $|x_n-x_m| < \varepsilon$ for all $n \ge m \ge p$. Now, \begin{align*} |x_n-x_m|=&|x_n - x_{n-1}+x_{n-1} -x_{n-2}+x_{n-2} + \cdots -x_m|\\ \leq & |x_n-x_{n-1}| + |x_{n-1}-x_{n-2}| + \cdots +|x_{m+1}-x_m|\\ \leq & (2/5)^{n-1} + (2/5)^{n-2} + \cdots +(2/5)^{m-1}\\ =& \dfrac{1-(2/5)^n}{1-\frac 25}\times (2/5)^m = \frac 53 (1-(2/5)^n) (2/5)^m \leq \frac 53 (2/5)^m \end{align*} so you just choose $p$ in such a way that $\frac 53 (2/5)^p < \varepsilon$.
If the limit exists, and we know it exists because Cauchy sequences in $\mathbb{R}$ are convergent, you have that $$ x_{n+1} = \frac{x_n^2+1}{5} \Rightarrow L=\frac{L^2+1}{5} $$ and so $L = \frac{1}{2} \left(5\pm \sqrt{21}\right)$. Knowing that $L \in [0,1]$, we must choose the "-" sign and conclude that $L=\frac{1}{2} \left(5-\sqrt{21}\right)$.