How does Lambert's W behave near ∞? How does $W$ behave near $+\infty$ compared to $\log$?  In particular, I'm interested in the asymptotic expansion of
$$\frac{W(x)}{\ln(x)}$$
near $\infty$ (but along the positive real line, if that matters).  Clearly $W(x)\sim\ln(x)$ as $x$ increases, and the next term looks hyperbolic.
 A: One of the original papers by Corless et al. on the Lambert function gives the following series expansion at $\infty$ for the (principal branch of the) Lambert function:
$$W_0(z)=\ln\;z-\ln\ln\;z+\sum_{n=1}^\infty\left(\frac{-1}{\ln\;z}\right)^n \sum_{m=1}^n (-1)^m \left[n\atop{n-m+1}\right]\frac{(\ln\ln\;z)^m}{m!}$$
where $\left[n\atop m\right]$ is a Stirling cycle number.
A: According to Theorem 2.7 of this paper, for every $x \geq e$,
$$
\log x - \log \log x + \frac{1}{2}\frac{{\log \log x}}{{\log x}} \le W(x) \le \log x - \log \log x + \frac{e}{{e - 1}}\frac{{\log \log x}}{{\log x}}
$$
(with equality only for $x=e$). Note that $\frac{e}{{e - 1}} \approx 1.582$.
EDIT (cf. Juan's answer). According to Wolfram MathWorld, an asymptotic formula which yields reasonably accurate results for sufficiently large $x$ is 
$$
W(x) = L_1  - L_2  + \frac{{L_2 }}{{L_1 }} + \frac{{L_2 ( - 2 + L_2 )}}{{2L_1^2 }} + \frac{{L_2 (6 - 9L_2  + 2L_2^2 )}}{{6L_1^3 }}
$$
$$
+ \frac{{L_2 ( - 12 + 36L_2  - 22L_2^2  + 3L_2^3 )}}{{12L_1^4 }} + \frac{{L_2 (60 - 300L_2  + 350L_2^2  - 125L_2^3  + 12L_2^4 )}}{{60L_1^5 }} + O\bigg[\bigg(\frac{{L_2 }}{{L_1 }}\bigg)^6 \bigg],
$$
where
$$
L_1 = \log x
$$
$$
L_2 = \log \log x.
$$
Useful references can be found in that link.
A: This Mathematica program takes a while to evaluate:
Clear[x, a, nn, b, z]
nn = 200;
z = 100
a = Series[Exp[-x], {x, N[Log[z], 500], nn}];
b = Normal[InverseSeries[Series[x/a, {x, 0, nn}]]];
x = z;
N[b, 20]
N[LambertW[x], 30]

It appears to give LambertW(100) = 3.3856301402900501849... but it uses 200 terms of the power series of Exp[-x] expanded at Log[100] to get there.
