Compute limn→∞sinsin…sinn \lim\limits_{n \to \infty }\sin \sin \dots\sin n

I need your help with evaluating this limit:

limnsinsinsinn compositionsn,

i.e. we apply the sin function n times.

Thank you.

Answer

The first sine is in I1=[1,1] hence the nth term of the sequence is in the interval In defined recursively by I1=[1,1] and In+1=sin(In). One sees that In=[xn,xn] where x1=1 and xn+1=sin(xn). The sine function is such that 0sin(x)x for every nonnegative x hence (xn) is nonincreasing and bounded below by zero hence it converges to a limit . The sine function is continuous hence =sin(). The only fixed point of the sine function is zero hence =0. This proves that xn0, that the sequence (In) is nonincreasing and that its intersection is reduced to the point zero and finally, that the sequence considered in the post converges to zero.

Edit: The argument above shows that for every sequence (zn), the sequence (sn) defined by sn=sinsinsin(zn) (the sine function being iterated n times to define sn) converges to zero. In other words, there is nothing particular about the choice zn=n.

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Source : Link , Question Author : Community , Answer Author : Did

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