How does Lambert's W behave near ∞?

Solution 1:

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.

Solution 2:

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.