How to prove the convergence of $u_{n+1}=\frac{\sin(u_n)}{n+1}$? I am first asked to prove that for all n, $u_n \in [0,1]$ for $u_0=1$. This is my attempt:
Let's first study the function defined as $f_x(n)=\frac{\sin(x)}{1+n}$ where $x$ is fixed and $x \in[0,\frac{2}{\pi}]$. We can immediately see that $0\leq f_x(n) \leq \frac{\pi}{2(n+1)}$.
We also know that $\sin$ is strictly decreasing when $x$ tends to $0$ on the interval [0,1]. Thus we can now say a few things:


*

*$f_x(n)$ is decreasing

*As $f_x(n) \leq \frac{\pi}{2(n+1)}$

*$\forall n \geq 1$ $\frac{\pi}{2(n+1)}\leq 1 $

*Thus $\sin(\frac{\pi}{2(n+1)})$ is decreasing 

*Thus $\sin(\frac{\pi}{2(n+1)}) \leq \sin(\frac{\pi}{2n}) $

*Thus $u_{n+1} \leq u_n$

*Finally as $u_0=1$ we have $u_n \leq 1$

*As $\sin(x) \geq 0 \forall x \in [0,\frac{\pi}{2}]$ we have $un>0$, thus $u_n \in [0,1]$ for all n.


I think my reasoning can be pretty confusing, but I am incapable of finding something clearer. My thoughts are pretty disorganised for such cases.
Now I need to show that $\sum u_n$ converges. But here I don't know how to proceed on doing this. 
 A: Here is a straightforward approach, if you want to have an answer following your approach, then leave a comment.
We prove inductively that $0≤u_n≤\frac{1}{n!}$ for all $n\in\mathbb{N}_0$. This is true for $n=0$, so assume it to be true for an $n\in\mathbb{N}_0$ fix. Then using that whenever $0≤x$ then $0≤\sin(x)≤x$ we obtain
$$
0≤u_{n+1}=\frac{\sin(u_n)}{n+1}≤\frac{u_n}{n+1}\le\frac{1}{(n+1)!}.
$$
Thus the inequality holds for all $n$. By the squeeze theorem we obtain that $\lim_{n\to\infty}u_n=0$ and by comparison we obtain the convergence of $\sum_{n=0}^{\infty}u_n$.
A: You have
$$\forall n\geq 0\; |u_{n+1}|\leq \frac{1}{n+1}$$
$$\implies \lim_{n\to+\infty}u_n=0$$
$$\implies \sin(u_n)\sim u_n \;(n\to+\infty)$$
$$\implies u_{n+1}\sim \frac{u_n}{n+1} .$$
by induction, you prove that $u_n>0$.
thus
$$\frac{u_{n+1}}{u_n}\sim \frac{1}{n}$$
and by ratio test, $\sum u_n$ converges.
A: Given that $u_0 = 1$, and 
$$u_{n+1} = \frac{ \sin u_n }{n+1} \ \mbox{ for } \ n = 0, 1, 2, 3, \ldots,$$
we note that $u_0 \in [0,1]$, and if $u_n \in [0,1]$, then $$0 \leq u_n \leq 1 < \frac{\pi}{2},$$ which implies that 
$$0 \leq \sin u_n < 1,$$
and hence $$0 \leq u_{n+1} < \frac{1}{n+1} \leq 1$$
for all $n = 0, 1, 2, 3, \ldots$. Here we have used the fact that the sine function is strictly increasing on $[0, \frac{\pi}{2}]$ and that $\sin 0 = 0$ and $\sin \frac{\pi}{2} = 1$. 
Now, using the auxiliary function $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(t) = t - \sin t$ for all $t \in \mathbb{R}$, we note that 
$$\frac{\mathrm{d} f(t)}{\mathrm{d} t} = 1- \cos t > 0$$ if $0 < t < 2 \pi$, showing that this function is strictly increasing on $[0, 1]$. Now $f(0) = 0$. So $f(t) \geq 0$ for all $t \in [0, 1]$. That is, $$t \geq \sin t \ \mbox{ for all } \ t \in [0, 1].$$
Thus, we obtain
$$u_{n+1} = \frac{\sin u_n}{n+1} \leq \frac{u_n}{n+1} \leq u_n$$
for all $n = 0, 1, 2, 3, \ldots$, thus showing that this sequence is monotonically decreasing. 
Thus we have a monotonically decreasing sequence that is also bounded below. Hence this sequence is convergent. 
Let $u$ denote the limit of this sequence. Then $u$ must satisfy 
$$0 \leq u \leq 1 \ \ \mbox{ and } \ \ u = \lim_{n \to \infty} u_{n+1},$$
and therefore $u$ must be $0$. 
