Let $f:\mathbb R\to \mathbb R$, $f(x)=x-\arctan(x)$.
Consider the sequence $(x_n)_{x\in\mathbb N}$, defined as $ x_0>0$ and $x_{n+1}=f(x_n)$. Prove that $x_n$ is bounded and calculate $\lim_{n\to \infty}x_n$.
I know that $f(x)$ is increasing on $\mathbb R$ and it's positive for $x>0$. That means $x_n$ is increasing and positive as well, since $x_0>0$. To prove convergence I know that I have to show that the series is $(1)$ bounded and $(2)$ monotone. $x_n$ is bounded below by $0$ and clearly increasing but I would think it has no upper bound? This clearly is wrong, what am I missing?
I appreciate any help!
Note that since $x>\arctan(x)$ for $x>0$ ($f(0)=0$ and you already noted that $f$ is increasing), it follows that for $x_0>0$ and $n\geq 0$, we have that $x_n>0$ and $$x_{n+1}=x_n-\arctan(x_n)< x_n$$ that is $(x_n)$ is positive and strictly decreasing. You can easily show that its limit is zero.