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 $f(x)<x$ 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 $0<a_n<1$ we get $L=0$.

For $x\in[-1,0)$ we have $f'(x)>1$ so again $f(x)<x$ 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 $f(x)<x$ 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 $\lim_{n\to\infty}\sqrt{n}a_n=\lim_{n\to\infty}0=0$

| cite | improve this answer | |

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

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

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – zare023 Dec 10 '19 at 1:00
  • $\begingroup$ Oh yes sorry I’ve just covered the case $a_0\ge 0$, I’m going to complete that. $\endgroup$ – user Dec 10 '19 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.