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
$$\lim _{n\rightarrow \infty}(n^3(x_{2n+1}-2\pi n- \frac{L}{n}))= L_2$$
?
 A: 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
$$
L_1=\frac{1}{2\pi}, \ L_2=\frac{-5}{6(2\pi)^3}=\frac{-5}{48\pi^3}
$$
A: Here is a solution for the first limit without expansions:
First we need to find some facts about the sequence $x_n$:


*

*note that in each interval $(2n\pi,(2n+1)\pi)$ we have two solutions $x_{2n+1}<x_{2n+2}$.

*therefore $x_{2n+1}=2\pi n +\varepsilon_n$ where $\varepsilon_n <\pi/2$. This proves at once that $\displaystyle \frac{x_{2n+1}}{n} \to 2\pi$

*furthermore we have $\sin \varepsilon_n =\sin(x_{2n+1})=\frac{1}{x_{2n+1}} \to 0$, and therefore $x_{2n+1}-2n \pi=\varepsilon_n \to 0$.
Now we are able to attack the first limit noting that $\sin y/y \to 1$ as $y \to 0$.
$$\lim_{n \to \infty} n(x_{2n+1}-2\pi n)= \lim_{n \to \infty} \frac{n}{x_{2n+1}}\cdot \frac{x_{2n+1}-2\pi n}{\sin(x_{2n+1}-2\pi n)}\cdot \frac{\sin x_{2n+1}}{\frac{1}{x_n}}=\frac{1}{2\pi}. $$
For the second limint, however, any method you choose will get you to write the expansion of $\sin$, because you need the higher order terms.
A: I'll given you a informal derivation of $L$, and leave it to you to formalize that. Since you need $\sin x = \frac{1}{n}$ for $x\sin x = 1$ to hold, the $x_n$ will generally lie close to some zero of $\sin x$, and get closer the larger $n$ gets. To figure out to which zero of $\sin x$ some $x_n$ lies close, look at the first few $x_n$. You get (a plot of $x\sin x$ helps!) $$
\begin{eqnarray}
  x_1 &\leq& \frac{\pi}{2} \\
  x_2 &\leq& \pi \\
  x_3 &\approx& 2\pi \\
  x_4 &\approx& 3\pi \\
  \ldots
\end{eqnarray}
$$
Thus, for large n, you have $x_n \approx (n-1)\pi$. If you were just interested in $\lim_{x\to\infty} (x_{2n+1} - 2\pi n)$, that approximation would suffice to see that the limit is zero!
But since you need $\lim_{x\to\infty} (n(x_{2n+1} - 2\pi n))$, we need to add a first-order term to our approximation of $x_n$. Since the limit contains only odd $x_n$, we're interested only in zeros of the form $2\pi n$. The slope of $\sin x$ at those zeros is +1 which yields the following first-order approximation $$
  \sin (2\pi n + \epsilon) \approx \epsilon
$$
From that, you get the following updated approximation of $x_n$ for odd n $$
  x_n \approx (n-1)\pi + \frac{1}{(n-1)\pi}
$$
Observe that if you compute $x_n\sin x_n$ (again, for odd n) with the approximation of $\sin x$ above, you get $((n-1)\pi + \frac{1}{(n-1)\pi})\frac{1}{(n-1)\pi)} \approx 1$. If you put that approximation into your limit, you get $$
  \lim_{n \to \infty} (n\frac{1}{2n\pi}) = \frac{1}{2\pi}
$$
For the second problem, you'll need to add third-order terms to those approximations.
