Prove that the recursively defined sequence is Cauchy. Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.
We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.
 A: If the sequence converges, then $\lim_{n \to \infty} x_n = L$ for some finite real number $L$. Then:
$$L(5+L) = 5+5L$$
$$5L+L^2 = 5+5L$$
$$L = \sqrt{5}$$
where we have discard $-\sqrt{5}$ as a solution because it is clear that the sequence $(x_n)$ consists only of elements greater than or equal to $0$. Now, we prove two things;

*

*$(x_n)$ is nondecreasing


*$(x_n)$ is bounded above by $\sqrt{5}$.
We prove the second statement by induction. So, clearly, the result holds when $n=1$. Suppose that it holds for arbitrary $n$. Then:
$$x_{n+1} = \frac{5+5x_n}{5+x_n} = 5 \cdot \frac{x_n+1}{x_n+5} = 5 \cdot \left(1 - \frac{4}{x_n+5} \right)$$
$$x_{n+1} = 5-\frac{20}{x_n+5}$$
Then, $x_n <  \sqrt{5} \implies x_n+5 < \sqrt{5}+5 \implies -\frac{1}{x_n+5} < -\frac{1}{5+\sqrt{5}}$. So:
$$x_{n+1} = 5-\frac{20}{x_n+5} < 5-\frac{20}{5+\sqrt{5}} = 5-\frac{20}{5^2-5} \cdot (5-\sqrt{5}) = 5-(5-\sqrt{5}) = \sqrt{5}$$
So, of course, it follows that $(x_n)$ is bounded above by $\sqrt{5}$.
We will now prove that it is nondecreasing. We have the following;
$$x_{n+1}-x_n = \frac{5+5x_n}{5+x_n}-x_n = \frac{5-x^2_n}{5+x_n}$$
However, we just proved that $(x_n)$ is bounded above by $\sqrt{5}$. So:
$$\forall n \in \mathbb{N}: 0 < x_n \leq \sqrt{5}$$
$$\forall n \in \mathbb{N}: x^2_n \leq 5$$
$$\forall n \in \mathbb{N}: 5-x^2_n \geq 0$$
which implies that $x_{n+1} \geq x_n$. By the Monotone Sequence Theorem, $(x_n)$ is convergent and, hence, is a Cauchy Sequence. $\Box$
I'm not entirely sure if there is a fast way to prove that this is Cauchy using only the definition. I suspect that you'd have to do quite a bit of work for it but I certainly haven't tried thinking about it too hard.
A: Clearly $x_n>0$. Noting
$$ x_{n+1}-\sqrt5=\frac{5+5x_n}{5+x_n}-\sqrt5=\frac{5+5x_n-5\sqrt5-\sqrt5x_n}{5+x_n} =\frac{5-\sqrt5}{5+x_n}(x_n-\sqrt5)$$
one has
$$ |x_{n+1}-\sqrt5| =\frac{5-\sqrt5}{5+x_n}|x_n-\sqrt5|\le\frac{5-\sqrt5}{5}|x_n-\sqrt5|.$$
This gives
$$ |x_n-\sqrt5|\le\bigg(\frac{5-\sqrt5}{5}\bigg)^{n-1}|x_1-\sqrt5|.$$
Thus
$$ \lim_{n\to\infty}|x_n-\sqrt5|=0 $$
or
$$ \lim_{n\to\infty}x_n=\sqrt5. $$
A: There is another shortcut solution, if you are allowed to use the Banach fixpoint theorem considering the function
$$f(x) = \frac{5+5x}{5+x} = 1+\frac{4x}{5+x}$$
Now, you see immediately that
$$x>0 \Rightarrow f(x) > 1 \text{ and } f(x) < 5$$
It follows that
$$f: [1,5] \rightarrow [1,5]$$
Since
$$f'(x) = \frac{20}{(5+x)^2}\Rightarrow |f'(x)| \leq \frac{20}{36}=\frac 59 < 1 \text{ on } [1,5]$$
it follows, that $f$ is a contraction on $[1,5]$ and hence $f$ has a uniquely determined fixpoint $f(x^{\star}) = x^{\star}$ and any recursion
$$x_{n+1} = f(x_n) \text{ with } x_1 \in [1,5]$$
will converge to this fixpoint.
So, for any starting point $x_1 \in [1,5]$ the recursion is a Cauchy sequence.
A: $$
x_{n+1} - \sqrt 5 = \frac{5+5x_n}{5+x_n} - \sqrt{5} = \frac{(5-\sqrt{5})(x_n-\sqrt 5)}{5+x_n}
$$
$$
x_{n+1} + \sqrt 5 = \frac{5+5x_n}{5+x_n} + \sqrt{5} = \frac{(5+\sqrt{5})(x_n+\sqrt 5)}{5+x_n}
$$
Therefore
$$
\frac{x_{n+1}-\sqrt 5}{x_{n+1}+\sqrt 5}=\frac{5-\sqrt 5}{5+\sqrt 5} \cdot \frac{x_n-\sqrt 5}{x_n+\sqrt 5}\\
\Rightarrow \frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{5-\sqrt 5}{5+\sqrt 5}\right)^{n-1} \cdot \frac{1-\sqrt{5}}{1+\sqrt{5}}
$$
then you can get a closed form solution for $x_n$ and the rest should be straightforward.
Please check this post: Recursive sequence depending on the parameter
I just learned about Möbius transformation the other day.
