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Let $a_1=2$ and $a_i=2\sin\frac{a_{i-1}}2$ for $i\ge2$. Prove that $\sum\limits_{i=1}^{2024}a_i<314$.

In fact, $\sum\limits_{i=1}^{2024}a_i\approx298.796$, so the inequality is very strong.

I tried to establish inequalities with Taylor's series, and got $\sum\limits_{i=1}^{2024}a_i<582$. This is not enough. We need better (more accurate) ways to estimate the series.

This question is from the "$\pi$ day math contest" of THU, which has ended.

Theo Bendit
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youthdoo
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    If you can prove the inequality $2\sin(\tfrac{\sqrt{\tfrac{12}{k}}}{2}) \le \sqrt{\tfrac{12}{k+1}}$, then, by induction $a_k \le \sqrt{\tfrac{12}{k}}$, from which it follows $\sum_{k = 1}^{2024}a_k \le \sum_{k = 1}^{2024}\sqrt{\tfrac{12}{k}} \le \sqrt{12}(2\sqrt{2024}-1) \approx 308.23$. – JimmyK4542 Mar 16 '24 at 05:16
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    What JimmyK4542 said! Writing $a_n=2b_n$ we have the recurrence $b_n=\sin(b_{n-1})$. Which reminds veterans of the site of this question. Potentially this is a duplicate, but the focus there is different, so we need to check a few things (and I want to solicit more opinions before acting anyway). – Jyrki Lahtonen Mar 16 '24 at 05:44
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    @JyrkiLahtonen: Two of the answers to that question prove the asymptotic behavior of the sequence, without explicit bounds. In the third answer it is shown that $\frac{3}{x^2} + 1 + \cdots<\frac{3}{\sin^2 x}$ and that is exactly the inequality from JimmyK4542's comment, unless I am mistaken. But the proof is not very rigorous, claiming that some “... remaining coefficients are positive.” – I would not close it as a duplicate. – Martin R Mar 16 '24 at 06:00
  • Thanks @MartinR. So I will refrain from voting to close as a dupe for now. A fond hope is that some of the other threads linked to that target may have the missing pieces (happens occasionally). Material to dig through to anyone so inclined :-) – Jyrki Lahtonen Mar 16 '24 at 06:11

2 Answers2

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The following is strongly inspired by JimmyK4542's comment and Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$.

Set $b_n = a_n/2$, then $b_1 = 1$ and $b_{n+1} = \sin(b_n)$. In Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ it is shown that $$ b_n \sim \sqrt{\frac{3}{n}} \text{ for } n \to \infty \, . $$ This suggests that an explicit estimate of the form $$ \tag{1} b_n \le \sqrt{\frac{3}{n+2}} $$ might hold for all $n \ge 1$. If that is true then $$ \begin{align} \sum_{n=1}^{2024} a_n &\le 2 \sum_{n=1}^{2024} \sqrt{\frac{3}{n+2}} = 2 \sqrt 3 \sum_{n=3}^{2026} \frac{1}{\sqrt n} \\ &< 2 \sqrt 3 \int_2^{2026} \frac{dx}{\sqrt x} = 4 \sqrt 3 (\sqrt{2026} - \sqrt 2) \\ &\approx 302.048 < 314 \, . \end{align} $$

In order to prove $(1)$ by induction we need to show that $$ \sin \sqrt{\frac 3k} \le \sqrt{\frac{3}{k+1}} \text{ for } k \ge 3 \, . $$ With the substitution $x = \sqrt{3/k}$ this is equivalent to $$ \tag{2} \sin^2(x) \le \frac{3x^2}{3 + x^2} \text{ for } 0 < x \le 1 $$ or $$ \tag{3} \frac{1}{\sin^2(x)} > \frac{1}{x^2} + \frac 13 \text{ for } 0 < x \le 1 \, . $$

I will now prove inequality $(3)$, but I wonder if there is a simpler way.

For $0 < x < \pi$ is (see e.g. here) $$ \frac{1}{\sin(x)} = \csc(x) =\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}2\left(2^{2n-1}-1\right)B_{2n}}{(2n)!}}x^{2n-1} \\ = \frac 1x + \frac{1}{6} x +\frac{7}{360}x^{3}+\frac{31}{15120}x^{5}+\cdots $$ where $B_n$ are the Bernoulli numbers. $B_{2n}$ is negative if $n$ is even, and positive otherwise, so that $(-1)^{n+1}B_{2n}$ is positive, i.e. all non-zero coefficients in that series are positive.

It follows that for $0 < x < \pi$ $$ \frac{1}{\sin^2(x)} > \left(\frac 1x + \frac x6\right)^2 > \frac{1}{x^2} + \frac 13 $$ and that completes the proof.

Martin R
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  • I don't understand how the form $\sqrt{\frac3{n+2}}$ was obtained, or at least predicted? – youthdoo Mar 16 '24 at 09:33
  • @youthdoo: I looked for an upper bound which is asymptotically $\sqrt{3/n}$ for large $n$. We have $b_1 = 1$, therefore I “guessed” that $b_n \le \sqrt{3/(n+2)}$ might do the job. Then I proved that this is indeed true. – Martin R Mar 16 '24 at 09:40
  • I see. This is difficult for a contest where people couldn't "look for large $n$". – youthdoo Mar 16 '24 at 09:50
  • Consider the function $$f(x)=\frac{1}{\sin^2(x)}-\frac{1}{x^2},\quad x\in(0,\pi),$$ we can get $$f'(x)=2\frac{\sin^3x-x^3\cos x}{x^3\sin^3x}>0,\forall x\in(0,\pi),$$ by $$\left(\frac{\sin x}{x} \right)^3 > \cos x,\forall x\in(0,\pi).$$ So $f(x)>f(0)=\lim_{x\to0}f(x)=\frac13$, for $x\in(0,\pi)$. – Riemann Mar 16 '24 at 09:58
  • @Riemann: Yes, that is a bit simpler. The last inequality is proven here: https://math.stackexchange.com/q/2088715/42969. – I am still looking for the “nicest” proof that $\frac{1}{\sin^2(x)} > \frac{1}{x^2} + \frac 13$, perhaps I will post that as a separate question some time ... – Martin R Mar 16 '24 at 10:05
  • @ Martin R That will be nice. – Riemann Mar 16 '24 at 10:14
  • I wonder what could be done using $\sin(x) <x, e^{-\frac{x^2}{6}}$ which is very tight. By the way, in $(2)$, isn't $x^2$ instead of $x^3$ ? – Claude Leibovici Feb 23 '25 at 04:53
  • @ClaudeLeibovici: Indeed, that was a typo. Strange that nobody noticed it for so long. Thanks! – Martin R Feb 23 '25 at 05:13
  • @ClaudeLeibovici: Yes, $\frac{x}{\sin x} > e^{x^2/6} > 1 + x^2/6$ for $0 < x < \pi$ is good enough for the desired estimate. – Martin R Feb 23 '25 at 05:20
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Starting from @Martin R's smart answer

$$ \sum_{n=1}^{p} a_n ~\le~ 2 \sum_{n=1}^{p} \sqrt{\frac{3}{n+2}}=2 \sqrt{3} \left(\zeta\left(\frac{1}{2},3\right)-\zeta\left(\frac{1}{2},p+3\right)\right)$$

If $p=2024$, the result is $300.912$.

Searching for the largest integer $p$ such that $\sum_{n=1}^{p} a_n < 314$ leads to $p=2197$ while the summation would give $p=2226$.

Much more complex would be to use the tighter bound $$\sin(x) <x\, e^{-\frac{x^2}{6}}$$ but it works.

Comparing the norms $$\Phi_1=\int_0^1\Bigg(\sin^2(x)-\frac{3 x^2}{3+x^2} \Bigg)^2\, dx=1.52696\times 10^{-4}$$ $$\Phi_2=\int_0^1\Bigg(\sin^2(x)-x^2\, e^{-\frac{x^2}{3}} \Bigg)^2\, dx=5.93439\times 10^{-6}$$

Claude Leibovici
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