Proving infinite intersection has only one point Given that:
$$ \begin{cases}a_1 = 1\\ a_{n+1} = \sqrt{a_n + 5} \end{cases} ~~~ \text{and} ~~~\begin{cases}b_1 = 4\\ b_{n+1} = \sqrt{b_n + 5} \end{cases}$$
Then:
$$ \bigcap_{n=1}^{\infty}[a_n,b_n] $$
Has only one point.
The thing is that I am stuck, it looks like the intervals are $$[1,4] \\ [\sqrt{5}, 3] \\ [ \sqrt{\sqrt{5} + 5}, \sqrt{8} ~] \\ \vdots $$
So it does go down (by going down I mean the length of the interval), and finally stops.. but where? How should I get the "final" point, is it just the limits?
$$ \lim a_n = \frac{1+\sqrt{21}}{2} \\ \lim b_n = \lim a_n$$
So the final point is $$ [ \lim a_n, \lim a_n]$$
I am really sure it is wrong, any help would be appreciated, Thanks!
 A: Yes, finding the limits of the two sequences is the way to go. Be sure to show that both limits do in fact exist, i.e. that both sequences converge. Also verify that this limit is contained in all intervals.
A: We have: $0 \le |b_n - a_n|= \dfrac{|b_{n-1} - a_{n-1}|}{\sqrt{b_{n-1}+5}+\sqrt{a_{n-1}+5}}< \dfrac{|b_{n-1} - a_{n-1}|}{2\sqrt{5}}<...< \dfrac{|b_1 - a_1|}{(2\sqrt{5})^{n-1}} = \dfrac{3}{(2\sqrt{5})^{n-1}}\implies \displaystyle \lim_{n \to \infty} |b_n - a_n| = 0$. Further, you can show: $[a_n, b_n] \subseteq [a_{n-1}, b_{n-1}]\implies \bigcap_{n=1}^{\infty}[a_n,b_n]= \{a\}= \{\frac{1+\sqrt{21}}{2}\}$ ( the common limit of both sequences ).
A: Suppose the sequence $a_{n}\to L$ or $b_{n}\to L$. Then $L$ must be nonnegative and real since $f(x)=\sqrt{x+5}$ has the property that $f([0,\infty))\subset [0,\infty)$. Moreover, $L$ must be a fixed point of the recursive equation. That is to say, if a solution of a recursive equation converges it must converge to an equilibrium solution.
$L=\sqrt{ L + 5 }$ so $L^{2}=L+5$, thus $L=\frac{1+\sqrt{21}}{2}$ is the unique non-negative equilibrium solution. Thus it suffices to show that $a_{n}$ and $b_{n}$ both converge.
The simplest way to prove convergence is through a well known result for first order difference equations. If you don't know it (here I mean can't use it in the exercise), the next easiest way is to observe that the sequences are monotone and bounded.
A: Have a look at the Banach fixed-point theorem; it guarantees existence and uniqueness of a fixed point of contractions.
You just need to show that $f(x) = \sqrt{x+5}$ is a contraction on $[1,4]$ which is not difficult.
