# About the sequence recursively defined by $x_{n+1}=b\left(1-\frac{b}{4x_n}\right)$

I don't know how to handle this:

Given $b > 0$, consider the following recursively defined sequence:

• $x_1=b$
• $x_{n+1}=b\left(1-\frac{b}{4x_n}\right)$, for all $n \ge 1$.

I've seen that the limit is $\frac{b}{2}$, but I don't know how to prove that it is bounded and decreasing. For example, when I'm proving that is bounded, I don't know what to do with the term $x_n$. Please, if someone could give me any hint, I'd appreciate it so much.

Thanks!

• $x_{n+1} < x_n \iff \dfrac{1}{x_{n+1}} > \dfrac{1}{x_n} \iff -\dfrac{b}{4x_{n+1}} < -\dfrac{b}{4x_n}$... Then see how to get to $x_{n+2}<x_{n+1}$? So the sequence is decreasing if and only if $x_2<x_1$. As far as bounded, show $x_n \ge b/4$ for all $n$ by induction. Jan 26 '14 at 5:38
• @DavidPeterson you should put that in an answer! Jan 26 '14 at 5:39
• David, thank you so much! Jan 26 '14 at 5:43
• @DavidPeterson, the problem in your comment is that the first inequality to the left is true iff $\;x_n>0\;\;\forall\,n\;$ ... Jan 26 '14 at 5:49
• That's "clear", @Abrahamlure...as all the question is. It must be proved. Jan 26 '14 at 6:15

So... one has $x_{n+1}=u(x_n)$ for every $n$, where $u:x\mapsto b\left(1-\frac1{4x}\right)$. It happens that:
• $u(x)\lt x$ for every $x\gt\frac{b}2$
• $u\left(\frac{b}2\right)=\frac{b}2$
• $u$ is increasing on $x\geqslant\frac{b}2$
• Thus, for every $x\gt\frac{b}2$, $x\gt u(x)\gt \frac{b}2$
This shows that, for every $x_1\gt\frac{b}2$, the sequence $(x_n)$ is decreasing, and that $x_n\gt\frac{b}2$ for every $n$. Thus $(x_n)$ converges to a limit $\ell$. To identify $\ell$, solve $\ell=u(\ell)$.
Suppose that, if the series is bounded by $L$, then $L=b\left(1-\frac{b}{4L}\right)$ must apply. Expand as a quadratic equation in $L$ and solve for $L$. Guess what you get.