$x_1:=1$ , and $x_{n+1}:=x_n+\cos(x_n) ,\forall n \in \mathbb N$ , does $(x_n)$ converge ? If $x_1:=1$ , and $x_{n+1}:=x_n+\cos(x_n) ,\forall n \in \mathbb N$ , then is it true that the sequence $(x_n)$ is convergent ? If it does , what is the limit ?
 A: Hint: Show that $f(x)=x+\cos(x)$ is increasing, that $x_2=1+\cos(1) \geq x_1$, and that $x_1<\frac{\pi}{2}$. Then show that $x_n$ is increasing, and $x_n\leq \frac{\pi}{2}$ for all $n$. 
A: Hint:
With induction you can prove that $1\leq x_{n}<\frac{1}{2}\pi\wedge x_{n}<x_{n+1}$
for each $n$. So the sequence is increasing and bounded.
A: If you have the contraction mapping principle, you can use it on $T(x) = x + \cos(x)$ on an appropriate interval containing $x = {\pi \over 2}$. 
Alternatively, you can show $|x_n - {\pi \over 2}|$ decreases geometrically for $n$ large enough... if $y_n = x_n - {\pi \over 2}$, your recursion becomes $y_{n+1} = y_n - \sin(y_n)$, and $\sin y_n \sim y_n$ for small $y_n$.
A: Similar to the ideas given in the other answers, one can show the result using the following two properties:


*

*$0\leq x \leq \pi/2$ $\Rightarrow$ $0 \leq x+ \cos(x) \leq \pi/2$.

*$cos(x)\geq 0$, for $x\in[0,\pi/2]$.


To prove the first point, use
$x + \cos(x) = x + \sin(\pi/2-x)$ and the fact $\sin(x) \leq x$ for $x \geq 0$.
The first property says that $x_n \in [0,\pi/2]$. Used with the first, the second one proves that $x_n$ is increasing and hence convergent.
