# Convergence of series with $a_{n + 1} = \sin{a_n}$

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.

• Aug 12 '20 at 16:45

Hint: $$\sin(1/n) > 1/(n+1)$$ for $$n \ge 1$$, and show by induction that $$a_n \ge 1/n$$.

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.

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.

• 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... Aug 28 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 $o(a_n^2)$. Aug 29 at 9:52