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Given $a_0=1948$ and $a_{n+1}=\text{sin}(a_n)$, Calculate $\lim_{n \rightarrow \infty} a_n$.

My thoughts:

$\text{sin}(1948)>0$, so my intuition tells me that it's decreasing and converges to it's infimum, and I'm having hard time proving it.

Any help appreciated.

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  • $\begingroup$ Since $\sin x\ge -1$, we must have $a_n\ge -1$ for all $a_n$. $\endgroup$ – Eclipse Sun Mar 7 '17 at 11:31
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First, note that $0<a_1<\pi$, since $0<\sin(1948)<\pi$.

Note that for any $0<x<\pi$, we have $0<\sin(x)<x$ and for any $x<\pi$, we have $\sin(x)<\pi$. Thus, if $0<a_n<\pi$, then $0<\sin(a_n)<\pi$, which shows $0<a_{n+1}<\pi$. This proves $a_n>\sin(a_n)=a_{n+1}$ thus $(a_n)$ is decreasing.

Also, $\sin(x)>0$ for $0<x<\pi$, so we know $0<a_n<\pi$ for all $n>1$.

Thus, $(a_n)$ is bounded and decreasing, thus, converging. Now call the limit $L$. We know $\lim a_n=L=\lim a_{n+1}$, or, $$L=\lim a_{n+1}=\lim \sin(a_n)=\sin(\lim a_n)=\sin(L)$$

thus, $L=0$.

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  • $\begingroup$ How can I prove that the sequence is decreasing? $\endgroup$ – Itay4 Mar 7 '17 at 11:40
  • $\begingroup$ I edited my answer in an attempt to clear that up. $\endgroup$ – vrugtehagel Mar 7 '17 at 11:44
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Can you show that $$\frac{1}{a_{n+1}^2}> \frac13+\frac1{a_n^2}$$

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$0$ is a fixed point of $\sin$. You could try to show that it is attractive in your case, using the critera for the derivative: $$ \lvert f'(x_0) \rvert < 1 $$ So $x_0 = a_1 = \sin(1948) = 0.21\dotsc < \pi/2 = 1.57\dotsc$ and $\lvert f'(x_0) \rvert = \lvert \cos(x_0)\rvert < 1$.

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