---
author:
- |
    Brian Y. Sun[^1]  and J. X. Meng\
    [College of Mathematics and System Science,]{}\
    [Xinjiang University, Urumqi 830046, P.R.China]{}
title: 'Proof of a Conjecture of Z.-W. Sun on Trigonometric Series[^2]'
---

[GBK]{}[song]{}

 Recently, Z. W. Sun introduced a sequence $(S_n)_{n\geq
0}$, where $S_n=\frac{\binom{6n}{3n}
\binom{3n}{n}}{2(2n+1)\binom{2n}{n}}$, and found one congruence and two convergent series on $S_n$ by . Furthermore, he proposed some related conjectures. In this paper, we first give analytic proofs of his two convergent series and then confirm one of his conjectures by invoking series expansions of $\sin(t\arcsin(x))$ and $\cos(t\arcsin(x)).$

[ Divisibility; Congruences; Trigonometric series; Convergent series]{} [ Primary 11B65; Secondary 05A10, 11A07.]{}

Introduction {#sec:introduction}
============

Throughout the paper, we let $\mathbb{R}$, $\mathbb{N}=\{0,1,2,3,\ldots\}$ and $\mathbb{N}^+=\mathbb{N}\backslash \{0\}$ denote the set of real numbers, natural numbers and positive integer numbers, respectively.

Recently, Z. W. Sun [@sun] considered the divisibilities of products and sums concerned binomial coefficients and central binomial coefficients. He also studied the divisibility of $\binom{6n}{3n}\binom{3n}{n}$ and obtained the following result, $$\label{sun-divisibility}
2(2n+1)\binom{2n}{n}\Big |\binom{6n}{3n}\binom{3n}{n}\,\,\text{for all}\,\, n\in \mathbb{N}^+,$$ where $\binom{m}{n}=\frac{m!}{n!(m-n)!}$ is a binomial coefficient. We call $\binom{2n}{n}$ the central binomial coefficient (cf. [@Sloane A000984]). The binomial coefficients ${\binom{6n}{3n}}$ is the sequence [@Sloane A066802] and ${\binom{3n}{n}}$ is the sequence [@Sloane A005809].

According to , Z. W. Sun [@sun] introduced the following integer sequence [@Sloane A176898]. $$S_n=\frac{\binom{6n}{3n}\binom{3n}{n}}
{2(2n+1){\binom{2n}{n}}},
\,\, \text{for} \,\,n\in \mathbb{N}^+.$$ Here we list the values of $S_1,S_2,\ldots,S_8$ as follows:

5, 231, 14568, 1062347, 84021990,\
7012604550, 607892634420, 54200780036595.

Z. W. Sun [@sun] proved that for any odd prime $p$, $$S_p\equiv15-30p+60p^2\pmod {p^3}.$$ Additionally, Guo [@guo] studied the sequence and proved that $3S_n\equiv 0~\pmod {2n+3}$, positively answering a question of Z. W. Sun [@sun]. By setting $S_0=\frac{1}{2}$ and employing , Z. W. Sun also obtained $$\label{mathematica-find-1}
\sum_{n\geq 0}S_nx^n=\frac{\sin(\frac{2}{3}\arcsin(6\sqrt{3x}))}{8\sqrt{3x}}\left(0<x\leq\frac{1}{108}\right),$$ and $$\label{mathematica-find-2}
\sum_{n\geq 0}\frac{S_n}{(2n+3)108^n}=\frac{27\sqrt{3}}{256}.$$ Particularly, setting $x=1/108$ in , we derive $$\label{convergent-bound}
\sum_{n=0}^\infty\frac{S_n}{108^n}=\frac{3\sqrt{3}}{8}.$$ Moreover, he proposed the following conjecture.

\[conjecture-1\] There exist positive integers $T_1,T_2,\ldots$ such that $$\label{eq-conjecture-1}
\sum_{k=0}^\infty S_kx^{2k+1}+\frac{1}{24}-\sum_{k=1}^{\infty}T_kx^{2k}
=\frac{\cos(\frac{2}{3}\arccos(6\sqrt{3}x))}{12}$$for all real $x$ with $|x|\leq \frac{1}{6\sqrt{3}}$. Also, $T_p\equiv-2\pmod p$ for any prime $p$.

In this paper, we first deduce formulas and by utilizing a series expansion in [@stanley]. Besides, we confirm Conjecture \[conjecture-1\] , i.e.,

\[theorem\] Conjecture \[conjecture-1\] is true.

Trigonometric Series {#sec:Trigonometric Series}
====================

Before proving Theorem \[theorem\], we need some formulas and series expansions on trigonometric functions. We first note that $$\label{arcsin+arccos}
\arcsin(x)+\arccos(x)=\frac{\pi}{2},$$ where $x\in [-1,1]$. What’s more, the basic fact $$\label{cos(a+b)}
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$ is needed.

Here we also use two trigonometric series in [@stanley Ex.  44, p.  51], namely, $$\label{stanley-series-1}
\sin(t \arcsin(x))=\sum_{n\geq 0}(-1)^nt\left(\prod_{i=0}^{n-1}(t^2-(2i+1)^2)\right)\frac{x^{2n+1}}{(2n+1)!},$$ and $$\label{stanley-series-2}
\cos(t \arcsin(x))=\sum_{n\geq 0}(-1)^n\left(\prod_{i=0}^{n-1}(t^2-(2i)^2)\right)\frac{x^{2n}}{(2n)!}.$$ With the formula in hand, it is not difficult to derive the identities and .

To begin with, we consider the uniform convergence of series . Due to Stirling [@PD], we hold the following approximate formula, which was called Stirling’s formula, $$\label{stirling-formula}\Gamma(\alpha)\approx \left(\frac{\alpha-1}{e}\right)^{\alpha-1}\sqrt{2\pi(\alpha-1)}
,\,\,\text{as}\,\,\alpha\rightarrow\infty,$$ where $\Gamma(\alpha)$ is the gamma function and is defined by $$\Gamma(\alpha)=\int_0^{+\infty}x^{\alpha-1}e^{-x}dx\,\, \mbox{for~}
\alpha>0.$$ For more information on this formula, one can consult [@PD]. By Stirling’s formula , we have $$\label{convergent-radius}
\rho=\overline{\lim_{n\rightarrow\infty}}\sqrt[n]{S_n}=108,\,\,\text{as}
\,\,n\rightarrow \infty.$$ Thus by Cauchy-Hadamard’s theorem and , the series $\sum_{n=0}^\infty S_nx^n$ is uniformly convergent for $0<x<\frac{1}{108}$. So, before we make operations on $\sum_{n=0}^\infty S_nx^n$, we designate $x\in(0,1/108)$ in order that all the following operations are well defined.

\[in-int\]For all $a,b\in \mathbb{R}$ and $bx\in[-1,1]$, we have $$\int x\sin(a\arcsin(bx))dx=\frac{\frac{\sin \left((a-2) \arcsin(b x)\right)}{a-2}-\frac{\sin \left((a+2) \arcsin(b x)\right)}{a+2}}{4 b^2}+C,$$ where $C$ is any constant.

Now we are in a position to consider formulas and .

-   As to , let $t=2/3$ and $x=6\sqrt{3x}$ in , we obtain $$\label{sn}
        \begin{split}
        \sin(\frac{2}{3} \arcsin(6\sqrt{3x}))&=\sum_{n\geq 0}\frac{2}{3}
        \left(\prod_{i=0}^{n-1}((2i+1)^2 -\frac{4}{9})\right)
        \frac{{(6\sqrt{3x})}^{2n+1}}{(2n+1)!}\\
        &=4\sqrt{3x}\sum_{n\geq 0}\frac{12^n}{(2n+1)!}
        \left(\prod_{i=0}^{n-1}(6i+1)(6i+5)\right)x^n\\
        &=8\sqrt{3x}\sum_{n\geq 0}S_nx^n,
        \end{split}$$ which is nothing but .

-   For , let $$\label{ff}
        f(x)=\sum_{n=0}^{\infty}\frac{S_n}{2n+3}
        x^{2n+3}.$$ Thanks to (i), we get $$f'(x)=\sum_{n=0}^\infty S_nx^{2n+2}=\frac{x\sin(\frac{2}{3}\arcsin(6x\sqrt{3}))}{8\sqrt{3}}.$$ From Theorem \[in-int\], it follows that $$\begin{aligned}
        f(x)&=\frac{1}{8\sqrt{3}}\int_{0}^xt\sin(\frac{2}{3}\arcsin(6t\sqrt{3}))dt\\
        &=\frac{\frac{3}{4} \sin \left(\frac{4}{3} \sin ^{-1}\left(6 \sqrt{3} x\right)\right)-\frac{3}{8} \sin \left(\frac{8}{3} \sin ^{-1}\left(6 \sqrt{3} x\right)\right)}{3456 \sqrt{3}},
        \end{aligned}$$ which indicates $$\label{value}
        f(1/6\sqrt{3})=\frac{1}{6144}.$$ Therefore, we can deduce identity by invoking and .

In addition, combining formulas and , it is not difficult to obtain the following result,

\[prop-1\] For all $t\in \mathbb{R}$ and $x\in [-1,1]$, we have $$\sin(\frac{\pi t}{2})\sin(t \arcsin(x))+\cos(\frac{\pi t}{2})\cos(t \arcsin(x))=\cos(t \arccos(x)).$$

With above results, we are ready to prove Theorem \[theorem\].

Proof of the Theorem \[theorem\] {#sec:Prove Conjecture}
================================

We shall give two lemmas before giving the proof of Theorem \[theorem\]. The first lemma is Fermat’s simple theorem [@fermat].

\[fermat-theorem\] If $a$ is any integer prime to $m$, and if $m$ is prime, then $a^{m-1}\equiv1\,\pmod m.$

\[lem-modp\] For any prime $p$, we have $$\frac{1}{p}{\binom{3p-2}{p-1}}\equiv -2\, \pmod p.$$

By applying $$\begin{aligned}
\frac{1}{p}{\binom{3p-2}{p-1}}
&=\frac{(3p-2)!}{p!(2p-1)!}\\
&=2\prod_{j=1}^{p-2}
\left(\frac{3p}{j+1}-1\right)\end{aligned}$$ and $$\begin{aligned}
\prod_{j=1}^{p-2}
\left(\frac{3p}{j+1}-1\right)&\equiv (-1)^{p-2}\\
&=-1\,\pmod p,\end{aligned}$$ the conclusion can be derived at once.

Now we can prove Theorem \[theorem\].

*Proof of Theorem \[theorem\].* Firstly, we need to point out that can be rewritten as follows: $$\cos(\frac{2}{3}\arccos(6x\sqrt{3}))=\sin(\frac{\pi}{3})\sin(\frac{2}{3}\arcsin(6x\sqrt{3}))
+\cos(\frac{\pi}{3})(1-24\sum_{k=1}^{\infty}T_kx^{2k}).$$ By Proposition \[prop-1\], for $k=1,2,3,\ldots$, if we can find $T_k$ such that $$1-24\sum_{k=1}^{\infty}T_kx^{2k}=\cos(\frac{2}{3}\arcsin(6x\sqrt{3})),$$ then we can prove the first part of Conjecture \[conjecture-1\].

By , we see that $$\begin{split}
\cos(\frac{2}{3}\arcsin(6x\sqrt{3}))&=-\sum_{n\geq 0}\frac{4}{9^n}\left(\prod_{i=1}^{n-1}(6i+2)(6i-2)\right)
\frac{(6x\sqrt{3})^{2n}}{(2n)!}\\
&=-\sum_{n\geq 0}\frac{16^n}{3n-1}{\binom{3n}{n}}x^{2n}\\
&=1-\sum_{n\geq 1}\frac{16^n}{3n-1}{\binom{3n}{n}}x^{2n}.
\end{split}$$ For integers $n\geq 1$, if we set $T_n=\frac{16^n}{24(3n-1)}{\binom{3n}{n}}$, thus obtaining the desired sequence $(T_n)_{n\geq 1}$ for Conjecture \[conjecture-1\].

It is clear that $$\frac{16^n}{24(3n-1)}{\binom{3n}{n}}
 =16^{n-1}\left(2{\binom{3n-2}{n-1}}
 -{\binom{3n-2}{n}}\right)$$ is an integer. One can refer to [@sun2] for details. In view of Lemmas \[fermat-theorem\] and \[lem-modp\], we can get $T_p\equiv-2 \,\pmod p$ for any prime $p$. This completes the proof of Theorem \[theorem\].

[**Acknowledgements.**]{} I would like to thank the referee for valuable comments and suggestions. This work was supported by NSFC (No. 11171283).

[99]{} J. Chernick, On Fermat’s simple theorem, [*Bull. Amer. Math. Soc.*]{} **45** (1939), no.4, 269–274. P. Diaconis and D. Freedman, An elementary proof of Stirling’s formula, [*Amer. Math. Monthly.*]{} (1986), 123–125.

V. J. W. Guo, Proof of two divisibility properties of binomial coefficients conjectured by Z. W. Sun, [*Electron. J. Combin.*]{} **21**(2014), no. 2, \#P2.54. N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, . R. P. Stanley, [*Enumerative combinatorics.*]{} Vol. I, Cambridge University Press, Cambridge, 1997. Z. W. Sun, Products and sums divisible by central binomial coefficients, [*Electron. J. Combin.*]{} **20** (2013), no. 1, \#P9. Z. W. Sun, On divisibility of binomial coefficients, [*J. Aust. Math. Soc.*]{} **93** (2012), 189–201.

[^1]: Corresponding author. [Email: brianys1984@126.com.]{}

[^2]: This work is supported by NSFC (No. 11171283).
