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Given the sequence $a_n \mspace{10mu},\mspace{10mu}a_{n+1}=\sin(a_n)\mspace{10mu} and \mspace{10mu} a_1=1. $

Find if the series $\sum_{k=1}^{\infty}a_k$ converges or diverges.

First I found that $a_n$ is monotone decreasing sequence. If $a_n\in[0;\pi/2]$ then $\sin(a_n)\leqslant a_n$. And $a_n\leq 0$.

$\lim_{n\rightarrow \infty}a_n=0.$

Ratio test gives $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=1$. More investigation of the series is required.

abcd123
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Hrackadont
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3 Answers3

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As you mentioned, $(a_n)$ decreases to $0$. Therefore, you have $$a_{n+1}^{-2} - a_n^{-2} = \sin(a_n)^{-2} - a_n^{-2} = \left(a_n - \frac{a_n^3}{6} + o(a_n^4)\right)^{-2} - a_n^{-2}$$ $$=a_n^{-2} \left[\left(1 - \frac{a_n^2}{6} + o(a_n^3)\right)^{-2} - 1 \right] = a_n^{-2} \left(1 + \frac{a_n^2}{3} + o(a_n^2) - 1 \right) = \frac{1}{3} + o(1)$$

So the sequence $a_{n+1}^{-2} - a_n^{-2}$ converges to $\frac{1}{3}$. You can now use Cesaro to see that $$a_n^{-2} \sim \frac{n}{3}$$

Therefore $$a_n \sim \sqrt{\frac{3}{n}}$$

so by comparison, the series diverges.

J. W. Tanner
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TheSilverDoe
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  • How do you get from the square-bracket to the next expression? I mean $$(1+x)^{-2} = 1 - 2x + O(x^2) , ,$$ so shouldn't it be $$(1-a_n^2/6 + O(a_n^4))^{-2} = 1 + a_n^2/3 + O(a_n^4) , ?$$ Doesn't change anything, but just wondering... – Diger Aug 28 '21 at 15:03
  • @Diger Your formulas are correct, but I think mine are also correct :) I used Taylor expansions with $o(...)$ (and not $O(...)$), which are, in that precise case, a little bit less precise. So when you expand $\left(1 - \frac{a_n^2}{6} + o(a_n^3)\right)^{-2}$, you get first the terms $1$ and $a_n^2/3$, and then, all the following terms are neglectible w.r.t. $a_n^2$, so you get a $o(a_n^2)$. – TheSilverDoe Aug 29 '21 at 09:52
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Hint: $\sin(1/n) > 1/(n+1)$ for $n \ge 1$, and show by induction that $a_n \ge 1/n$.

Robert Israel
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You can show that $(u_n)$ is decreasing by proving that $a_n\in\left[0,\frac{\pi}{2}\right]$ for all $n\in\mathbb{N}$ and using $|\sin(x)|\leqslant |x|$, thus $(u_n)$ converges, and the limit is $0$ since $0$ is the only solution to $\sin\ell=\ell$. Let $\alpha\in\mathbb{R}$, then $$ a_{n+1}^{\alpha}=\left(a_n-\frac{a_n^3}{6}+\mathcal{O}(a_n^5)\right)^{\alpha}=a_n^{\alpha}\left(1-\frac{\alpha}{6}a_n^2+\mathcal{O}(a_n^4)\right) =a_n^{\alpha}-\frac{\alpha}{6}a_n^{2+\alpha}+\mathcal{O}(a_n^{4+\alpha})$$ Taking $\alpha=-2$ gives you that $\displaystyle\lim\limits_{n\rightarrow +\infty}\frac{1}{a_{n+1}^2}-\frac{1}{a_n^2}=\frac{1}{3}$ and, by Cesaro's theorem, $\displaystyle\lim\limits_{n\rightarrow +\infty}\frac{1}{n a_n^2}=\frac{1}{3}$. Thus $a_n\underset{n\rightarrow +\infty}{\sim}\sqrt{\frac{3}{n}}$ and the series $\sum a_n$ diverges.

Tuvasbien
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