Limit of a sequence

Find , if it exists, the limit of the sequence:

$$\mathop{\lim}\limits_{n\rightarrow{+\infty}}{({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n-\text{times}}})({n})}$$

Solution

The sequence \(u_{n}=({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n -{\text{times}}}})({1})\), \(n\in\mathbb{N}\), is strictly descending and bounded from below. So it converges to a real number \(\ell\) . But then \[\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{u_{n}}=\mathop{\lim}\limits_{n\rightarrow{+\infty}}{\sin({u_{n-1}})}\quad\Rightarrow\quad{\ell}=\sin{\ell}\quad\Rightarrow\quad{\ell=0\,.}\] So  \(\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{u_{n}}=0\) .
 
The figure below shows the terms of sequences \(\{{u_ {n}}\}_{u=1}^{\infty}\) and \(\{{-u_ {n}}\}_{u=1}^{\infty}\) on the \( y \) -axis  as  (oriented) arcs on the trigonometric unit circle.

Let the sequence \[a_{n}(y)=({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n-{\text{times}}}})({y})\, , \;n\in\mathbb{N}\, ,  \;y\in[{-1,1}]\,.\] Substituting  \(y=\sin({x})\) , where  \(x\in\mathbb{R}\), we have \(-1\leqslant\sin({x})=y\leqslant1\) . Because the function \(\sin\) is strictly increasing on  \([{-1,1}]\) by repeated implementation we have
\[\begin{array}{ccc}
 -u_1=& \sin(-1)\leqslant\sin({\sin(x)})=a_1(y)\leqslant\sin(1)= & u_1\\
 -u_2=& \sin({\sin(-1)})\leqslant\sin({\sin({\sin(x)})})=a_2(y)\leqslant\sin({\sin(1)})=& u_2\\
....&..................................................................................& .... \\
{-u_{n}}= & ({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n-{\text{times}}}})({-1})\leqslant({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n+1 -{\text{times}}}})({x})=a_{n}(y)\leqslant({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n-{\text{times}}}})({1})= & u_{n}\\
....&..................................................................................& ....
\end{array}\]
So   \(\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{a_{n}}(y)=0\), \(y\in[{-1,1}]\) and because  \(\sin({n+1})\in[{-1,1}]\), for every   \(n\in\mathbb{N}\), we have
\begin{alignat*}{2}
\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{a_{n}}(y)&=\mathop{\lim}\limits_{n\rightarrow{+\infty}}{a_{n}}(\sin({n+1}))=\mathop{\lim}\limits_{n\rightarrow{+\infty}}{({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n+1 \ -{\text{times}}}})({n+1})}\\
& =\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{({\underbrace{\sin\circ\sin\circ\ldots\circ\sin}_{n-{\text{times}}}})({n})}=0\,.
\end{alignat*}

The exercise can also be found in mathimatikoi.org