Limits of the solutions to $x\sin x = 1$ Let $x_n$ be the sequence of increasing solutions to $x\sin{x} = 1$. Define $$a = \lim_{n \to \infty} n(x_{2n+1} - 2\pi n) $$
and $$b = \lim_{n \to \infty} n^3 \left( x_{2n+1} - 2\pi n - \frac{a}{n} \right) $$
These limits can be evaluated, but they appear to be the first limits in some sequence. How can we find a general formula for all the limits in this sequence? 
(For example,  $$c = \lim_{n \to \infty} n^5\left( x_{2n + 1} - 2\pi n - \frac{a}{n} - \frac{b}{n^3} \right) $$ might be the next in the sequence). 
 A: Because you are always evaluating the limit, this is an asymptotic expansion of the explicit expression for the solutions. Write
$$x=2\pi n +\epsilon$$
You get
$$\sin \epsilon=\frac{1}{2\pi n +\epsilon}$$
Your first limit in this notation is
$$a=\lim_{n\to\infty}n\epsilon$$
We are seeking the series for $\epsilon$ expanded in inverse powers of $n$.
$$z=\frac{1}{2\pi n} = \frac{1}{\frac{1}{\sin \epsilon}-\epsilon}=f(\epsilon)$$
Formally, this is done. Use the Lagrange Inversion Theorem to express the inverse of $f(\epsilon)$ as a power series of $z$:
$$\epsilon(z)=\sum \frac{z^k}{k!} \lim_{x\to 0}\frac{d^{k-1}}{dx^{k-1}}\left(\frac{x}{f(x)}\right)^k$$
$$=\sum \frac{z^k}{k!} \lim_{x\to 0}\frac{d^{k-1}}{dx^{k-1}}\left(\frac{x}{\sin x}-x^2\right)^k$$
The first term (k=1):
$$\lim_{x\to 0}\left(\frac{x}{\sin x}-x^2\right)=1$$
The second term (k=2):
$$\lim_{x\to 0}\frac{d}{dx}\left(\frac{x}{\sin x}-x^2\right)^2=$$
$$\lim_{x\to 0}2\left(\frac{x}{\sin x}-x^2\right)\left(\frac{\sin x-x \cos x}{\sin^2 x}-2x\right)=0$$
Mathematica says that the third term is $-5$ and fourth is again zero. You can write
$$\epsilon=\frac{1}{2\pi n}-\frac{5}{3!}\frac{1}{(2\pi n)^3}+\frac{169}{5!}\frac{1}{(2\pi n)^5}-\frac{15063}{7!}\frac{1}{(2\pi n)^7}\cdots$$
I realize that this is not a very elegant solution and the terms in expansion don't seem to follow any simple pattern - the function $f$ is too ugly. I did half of this on paper so if I made any stupid mistakes, please point them out.
