Convergence of a sequence defined by recursion Let $x_1\in(0,\pi/2]$ and
$$x_{n+1}=x_n-\frac{1}{n+1}\sin x_n,\ \forall n\in\mathbb{Z}_{\geq1}$$
It could be seen that the sequence $\{x_n\}_{n=1}^{\infty}$ is positive and strictly decreasing, so it has a limit $\lim_{n\to\infty}x_n=c\geq0$. And then I could calculate that $c=0$ from this point of view:
$$
x_{n}-x_1=\sum_{k=2}^n(x_k-x_{k-1})=-\sum_{k=2}^n\frac{1}{k}\sin x_{k-1}<-\sin c\sum_{k=2}^nk^{-1}
$$
Because of the divergence of the harmonic series and the convergence of $\{x_n\}_{n=1}^{\infty}$, we must have $c=0$.
And my question raised during I did the derivation above: How quickly does the sequence $\{x_n\}_{n=1}^{\infty}$ converge? For example, will it satisfy $x_n\sim O(1/n)$? (i.e. Does the limit $\lim_{n\to\infty}nx_n$ exist?) Thanks for your help!
 A: On the region of interest, $\sin(x) \geq x - \frac{x^3}{6}$, and this approximation is extremely good for $x \ll 1$. Since it's easily seen that $0 \leq x_2 \leq 0.3006$, this approximation is basically exact if we start applying at at $x_2$.
Applying this approximation to the recurrence gives
$$
x_{n+1} 
\leq 
\frac{n}{n+1} x_n \left ( 1 + \frac{x_n^2}{6n} \right )
\leq
\frac{n}{n+1} x_n \left ( 1 + \frac{1}{6n} \right )
$$
The estimate $x_n^2 \leq 1$ is very bad. In fact, $x_n^2 \approx 0$. But this is good enough to show (by induction) that
$$
x_{n+1} 
\leq 
\frac{x_0}{n+1} \prod_{i = 1}^n \left ( 1 + \frac{1}{6i} \right )
$$
Applying our favorite estimate $1+x \leq e^x$, we see
$$x_{n+1} \leq \frac{x_0}{n+1} e^{\frac{1}{6} H_n}$$
where $H_n = \sum_{i = 1}^n \frac{1}{i}$ is the $n$th harmonic number.
Then since $H_n \leq \ln(n) + 1$, we see
$$x_{n+1} \leq \frac{x_0}{n+1} e^{\frac{1}{6} (\ln(n) + 1)}$$
And lastly
$$x_{n+1} \leq \frac{x_0}{n+1} n^{\frac{1}{6}}e^{\frac{1}{6}}$$
So $x_n = O(n^{-5/6})$. This might be good enough, but we can do better. After all, the approximation $x_n^2 \leq 1$ isn't very good.

Now that we know $x_n = O(n^{-5/6})$, we can write
$$x_{n+1} \leq \frac{n}{n+1} x_n \left ( 1 + O(n^{-8/3}) \right )$$
Then as before
$$x_{n+1} \leq \frac{x_0}{n+1} \exp \left ( \sum_{i=1}^n O(i^{-8/3}) \right )$$
We can upper bound this sum by the infinite version of it, which converges, so we have
$$x_{n+1} \leq \frac{x_0}{n+1} e^{O(1)}.$$
This gives $x_n = O \left ( \frac{1}{n} \right )$.
As a (fun?) exercise:
Do you see why iterating this idea again won't improve our big Oh past
$O \left ( \frac{1}{n} \right )$?

Lastly, with sage one can compute the constant $k$ (depending on $x_0$) so that $x_n \approx \frac{k x_0}{n}$.
For $x_0 = \frac{\pi}{2}$, for instance, we see $k \approx 0.387$. Then

*

*$x_{1000}                     = 0.000608664796133123$

*$\frac{k \frac{\pi}{2}}{1000} = 0.000608664342810232$
which is quite a good approximation.

I hope this helps ^_^
