1
$\begingroup$

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.

$\endgroup$
1
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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... $\endgroup$
    – Diger
    Aug 28 at 15:03
  • $\begingroup$ @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 $o(a_n^2)$. $\endgroup$ Aug 29 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.