<s>[INST] Sequence of solutions to $x\sin x=1$ 
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed. 

Consider a sequence $x_n, n\ge1$ formed by positive solutions to $x \sin{x}=1$.
How can we find 
$$\lim _{n\rightarrow \infty}(n(x_{2n+1}-2\pi n))= L$$
and
$$\lim _{n\rightarrow \infty}(n^3(x_{2n+1}-2\pi n- \frac{L}{n}))= L_2$$
?
 [/INST] Let $y$ be a variable tending to zero. Put 
$$
\begin{array}{l}
a=y-\frac{5}{6}y^3, \\ 
b=y-\frac{5}{6}y^3+\frac{169}{120}y^5
\end{array}
$$
Using Taylor expansions,  one finds that 
$$
\begin{array}{l}
\sin(a)\bigg(\frac{1}{y}+a\bigg)=1-\frac{169}{120}y^4+O(y^5), \\
\sin(b)\bigg(\frac{1}{y}+b\bigg)=1+\frac{5021}{1680}y^6+O(y^7)
\end{array}
$$
so for small enough $y$ there will be a $c\in ]a,b[$ such that $\sin(c)\bigg(\frac{1}{y}+c\bigg)=1$. This $c$ will satisfy $c=y-\frac{5}{6}y^3+O(y^5)$.
Applying this to $y=\frac{1}{2\pi n}$ yields
$$
L_1=\frac{1}{2\pi}, \ L_2=\frac{-5}{6(2\pi)^3}=\frac{-5}{48\pi^3}
$$
</s>