# Examine the convergence of the series $a_{n+1}=a_n-\arcsin(\sin^2a_n)$, where $a_o\in[-1,1]$, and find $\lim_{n\to\infty}\sqrt{n}a_n$.

My attempt:

Since $$0\leq\sin^2a_n<1$$ which means that $$0\leq\arcsin(\sin^2a_n)<\frac{\pi}{2}$$, so that the series will decrease when $$a_0\in[-1,1]\setminus\{0\}$$ without a limit, and that it will stay at $$0$$ when $$a_0=0$$.

If i am right (and I highly suspect that I'm not) that the sequence diverges, how would i find $$\lim_{n\to\infty}\sqrt{n}a_n$$? Should I divide the problem into different cases? If so, I've tried using Stolz-Cesàro when $$a_0=0$$, but that doesn't seem to lead anywhere.

Let $$f(x)=x-\arcsin(\sin^2(x))$$ then $$f'(x)=1-\frac{2\sin(x)\cos(x)}{\sqrt{1-\sin^4(x)}}$$ so $$f'(x)<1$$ for $$x\in(0,1]$$. Therefore $$f(x) for $$x\in(0,1]$$ since $$f(0)=0$$ so your sequence decreases. Also $$f(x)>0$$ since $$f(1)>0$$ and $$f$$ has a local maximum on $$(0,1)$$. So your sequence is decreasing and bounded and therefore convergent. The limit $$L$$ must satisfy $$L=L-\arcsin(\sin^2(L))\Longleftrightarrow L=\pi z$$ with $$z\in\mathbb{Z}$$ and since $$0 we get $$L=0$$.

For $$x\in[-1,0)$$ we have $$f'(x)>1$$ so again $$f(x) for $$x\in[-1,0)$$. This also holds for $$x\in(-\pi,0)$$: Since $$f$$ is increasing on $$(-\frac{\pi}{2},0)$$ and has a local maximum on $$(-\pi,-\frac{\pi}{2})$$ $$f$$ must have a local minimum at $$(-\frac{\pi}{2}\mid-\pi)$$ since the function is decreasing before, increasing after and $$f'(-\frac{\pi}{2})$$ is undefined. Since $$f(x)=x\Longleftrightarrow x=\pi z$$ with $$z\in\mathbb{Z}$$ we have $$f(x) for $$x\in(-\pi,0)$$ and since $$f(x)=-\pi\Longleftrightarrow x=-\pi\vee x=-\frac{\pi}{2}$$ we have $$f(x)\geq-\pi$$ for $$x\in[-\pi,0]$$. So your sequence is decreasing and bounded and therefore convergent again. The limit $$L$$ must satisfy $$L=\pi z$$ again and since $$-\pi\leq a_n<0$$ we get $$L=-\pi$$.

For $$a_0=0$$ you don't need Stolz-Cesaro since $$a_n=0\forall n\in\mathbb{N}$$ so $$\lim_{n\to\infty}\sqrt{n}a_n=\lim_{n\to\infty}0=0$$

For the first part we have that if $$a_0=0 \implies a_n=0$$ and $$\sqrt n a_n =0$$.

For $$a_0 > 0$$

• $$a_{n}>0$$ (by induction)

• $$a_{n+1}< a_n \iff a_{n+1}- a_n= -\arcsin(\sin^2a_n) < 0$$

• $$a_n \to L$$ (monotone sequence theorem)

• $$L=L -\arcsin(\sin^2L) \implies L=0$$

For $$a_0 < 0$$ let consider $$b_n=\pi+a_n>0$$ then

$$a_{n+1}=a_n-\arcsin(\sin^2a_n) \iff b_{n+1}=b_n-\arcsin(\sin^2b_n)$$

• $$b_{n}>0$$ (by induction)

• $$b_{n+1}< b_n \iff b_{n+1}- b_n= -\arcsin(\sin^2b_n) < 0$$

• $$b_n \to L$$ (monotone sequence theorem)

• $$L=L -\arcsin(\sin^2L) \implies L=0$$

that is $$a_n \to -\pi$$.

For the second part, for the case $$a_0>0$$, let consider $$na_n^2$$ and by Stolz-Cesaro we have

$$na_n^2=\frac n{\frac1{a_n^2}} \implies \frac{n+1-n}{\frac1{a_{n+1}^2}-\frac1{a_n^2}}=\frac{a_{n+1}^2a_n^2}{a_n^2-a_{n+1}^2}\sim \frac{(a_n-a_n^2)^2a_n^2}{a_n^2-a_n+a_n^2}=\frac{(a_n-a_n^2)a_n}{2a_n-1} \to 0$$

and therefore $$\sqrt n a_n \to 0$$.

By a rough evaluation we can also claim that $$a_n\sim \frac 1n$$ indeed by $$a_n\sim cn^\alpha$$

$$c(n+1)^\alpha= cn^\alpha-\arcsin(\sin^2a_n)\sim cn^\alpha-c^2n^{2\alpha}$$

$$\left(1+\frac1n\right)^\alpha -1 \sim-cn^{\alpha} \implies 1+\frac{\alpha}n -1 \sim-cn^{\alpha} \implies c=1,\,\alpha=-1$$

indeed again by Stolz-Cesaro we have

$$na_n=\frac n{\frac1{a_n}} \implies \frac{n+1-n}{\frac1{a_{n+1}}-\frac1{a_n}}=\frac{a_{n+1}a_n}{a_n-a_{n+1}}\sim \frac{(a_n-a_n^2)a_n}{a_n-a_n+a_n^2}=1-a_n \to 1$$

• But doesn't the first part only hold for $a_0\geq0$? For example, if we plug in $-1$, we'll get about $-1.7$ and decreasing? – zare023 Dec 10 '19 at 1:00
• Oh yes sorry I’ve just covered the case $a_0\ge 0$, I’m going to complete that. – user Dec 10 '19 at 7:10