It's given that $a \in \mathbb{R}$ with $a > 0$ and $0<x_1<\frac{1}{a}$. The sequence $(x_n)_{n\in\mathbb{N}}$ is defined by $x_{n+1} = 2x_n-ax^2_n.$ Now, I first proved by induction that $x_n < \frac{1}{n}$ for all $n \in \mathbb{N}$. Then I was wondering, if it is enough to just show that $x_{n+1} = 2x_n-ax^2_n > 2x_n – \frac{1}{x_n}x^2_n = 2x_n-x_n = x_n$ to fully prove that $ x_{n+1} > x_n$ for all $n \in \mathbb{N}$ and therefore $(x_n)_{n\in\mathbb{N}}$ is monotonically increasing? Thanks for any corrections or tips, I'm just not sure if I might be missing something.
Show that a sequence is monotonically increasing.
calculusinductionsequences-and-series
Best Answer
You have to justify the inequality $x_n(2-ax_n) <\frac 1 a (2-\frac 1 a a)$. To prove this consider the function $f(t)=t(2-at)$ defined on $(0,\frac 1 a)$. Since $f'(t)=2-2at>0$ for all $t$ in the domain of $f$ this function is increasing. Since $t(2-at)=\frac 1 a$ when $t=\frac 1 a$ it follows (by looking at the extension of $f$ to $(0,\frac 1 a]$) that $f(t) <\frac 1 a$ for all $t$. So if we assume that $x_n <\frac 1 a$ we can take $t=x_n$ to see that $x_{n+1}<\frac 1 a$.