4

Let sequence $\{a_n\}$ be defined by $a_1\in\left(0,\sqrt3\right)$ and $a_{n+1}=a_n-\frac{\sin a_n}{n+1}$ for $n\ge1$. Prove that $na_n<\frac{3a_1}{3-a_1^2}$.

I think this question resembles this previous question of mine. Thus, I decided to use differential equations to estimate $a_n$. Generalize $\{a_n\}$ to $f(x)$, then $$f(x+1)-f(x)=-\frac{\sin f(x)}{x+1}$$ we can get an approximate equation $$f'(x)=-\frac{\sin f(x)}{x+1}.$$ Solving gives $f(x)=2\arctan\left(\frac c{x+1}\right)$. I checked that this does not help because $xf(x)\to2c$ as $x\to+\infty$, but this upper bound is not enough.

How to prove this inequality then?

youthdoo
  • 3,663
  • From the difference equation (your first equation with $f(x+1)-f(x)$), you can take the sum of all values from $0$ to $n$, which will be a telescopic sum for the left side, so you can get a closed form for $a_n$ in terms of a finite sum. Maybe that helps? – Kraken Feb 24 '25 at 05:39
  • @Kraken I tried telescoping series, and in many ways, but I did not succeed. – youthdoo Feb 24 '25 at 08:53
  • This is an interesting question. My numerical experiments indicate that the bound $na_n<\frac{3a_1}{3-a_1^2}$ is far from being tight. Can you tell something about the origin of this problem? – Martin R Mar 15 '25 at 09:28
  • @MartinR It is an Olympiad training problem. I know nothing more than this. – youthdoo Mar 15 '25 at 13:17

0 Answers0