1. ## A Problem about series

Let's consider a series defined by $\displaystyle x_{n+1} = sin(x_n)$. $\displaystyle x_0\in(0,pi)$
then construct a series : $\displaystyle \sum{x_n}^p$, n from 0 to $\displaystyle \infty$
Now discuss the convergence of the above series.

2. Originally Posted by ThePerfectHacker
Thus, $\displaystyle |x_n|\leq x_1 < 1$ but since $\displaystyle x_1^{1/n} \to 0$ it means $\displaystyle |x_n|^{1/n} \to 0$ by the root test it means the series converges.
Umm, unfortunately $\displaystyle x_1^{1/n} \to 1$, not 0, as $\displaystyle n\to\infty$.

I haven't been able to think of any way of attacking this problem. It's clear that $\displaystyle x_n\to 0$ as $\displaystyle n\to\infty$, but it approaches 0 so desperately slowly that I very much doubt whether $\displaystyle \sum x_n$ converges. Maybe the series of p'th powers will converge if p is large enough, but I don't see why even that should hold.

It's a very interesting problem! I hope someone can find a way to tackle it.

3. Originally Posted by Opalg
Umm, unfortunately $\displaystyle x_1^{1/n} \to 1$, not 0, as $\displaystyle n\to\infty$.
Sorry for the mistake.
(My approach would be to ignore the $\displaystyle p$ and try to do the simplified problem first.)

4. I can see now how to do this for p=1.

Since x_0 > 0 there is an integer k such that $\displaystyle x_0 > 1/k$. The strategy is to prove by induction that $\displaystyle x_n >1/(k+n)$. It will then follow (by comparison with the harmonic series) that $\displaystyle \sum x_n$ diverges.

The proof relies on the fact that $\displaystyle \textstyle \sin x > x-\frac16 x^3 = x\left(1-\frac16x^2\right)$ (for 0<x<1). This is because the power series for sin(x) is an alternating series with decreasing terms, so the partial sums are alternately greater than, and less than, the whole sum.

So suppose that $\displaystyle x_n>1/(k+n)$. Then $\displaystyle x_{n+1} = \sin(x_n) > \sin(1/(k+n)) > \frac1{k+n}\left(1-\frac1{6(k+n)^2}\right).$ To complete the inductive step, we want to show that this is greater than $\displaystyle \frac1{k+n+1} = \frac1{k+n}\left(1-\frac1{k+n+1}\right).$ But that is equivalent to $\displaystyle 6(k+n)^2 > k+n+1$, which is certainly true.

A slightly more elaborate form of this same argument shows that the series $\displaystyle \sum x_n^p$ diverges whenever p<2. I still have no idea what happens when $\displaystyle p \geqslant2$.