Let $a_0=1$ and $a_{n} = \sin(a_{n-1})$. Does the following series $$ \sum\limits_{n=0}^{\infty}a_n $$ converges?
I have no ideas...
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3Various postings, such as this, show that $a_n$ is asymptotically $\sqrt{3/n}$. Given this, what can you say about the sum? – Sangchul Lee May 14 '18 at 10:54
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1@SangchulLee, I can say that the sum is diverges) – Mikhail Goltvanitsa May 14 '18 at 13:53
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$a_n \geq\frac{1}{n} (n \geq2)$, because
if $a_{n-1} \geq \frac{1}{n-1},$ then $ a_n = \sin(a_{n-1}) \geq \sin(\frac{1}{n-1}) \geq \frac{1}{n}$
reason of $\sin(\frac{1}{n-1} ) \geq \frac{1}{n} (n\geq2)$ (graph)
so $\sum a_n \geq \sum \frac{1}{n}=\infty$
Sintist
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