TITLE: Asymptotics of unrooted labeled forests
QUESTION [5 upvotes]: It is well known that the number of unrooted labeled trees on vertex set
$[n]={1,2,...,n}$ is $n^{n-2}$. Let $U(z)$ be the exponential generating function of the sequence of these numbers. Then $F(z)=\text{exp}(U(z))$ is the exponential generating function for the number of unrooted forests of $[n]$. 
My question is, how do you prove that $n![z^n]F(z) \sim e^{1/2} n^{n-2}$. 
In other words, that is the asymptotic  number of unrooted forests on $[n]$. I found this result on page 406 of the Flajolet-Sedgewick book, without proof.   
If that helps, $U(z)= T(z) - T(z)^2 /2$ , where $T(z)$ is the exponential generating function for the numbers of rooted trees. But while $T(z)$ satisfies the relation $T=z \ \text{exp}(T)$, I could not find such a relation for F.

REPLY [6 votes]: The idea in the book by Flajolet-Sedgewick is to use singularity analysis. The generating function $F(z)=\exp(U(z))$ with $U=T-T^2/2$ inherits the dominant singularity $e^{-1}$ of $T(z)=-W(-z)$, where $W$ is the Lambert $W$ function. From the local expansion of $T(z)$
$$T(z)=1-\sqrt{2}\sqrt{1-ze}+\frac{2}{3}(1-ze)-\frac{11\sqrt{2}}{36}(1-ze)^{3/2}+O((1-ze)^2),\quad z\rightarrow e^{-1}$$
follows that of $F$:
$$F(z)=e^{1/2}\left(1-(1-ze)+\frac{2\sqrt{2}}{3}(1-ze)^{3/2}+O((1-ze)^2)\right).$$
From there, singularity analysis is straightforward. The asymptotic behaviour of the coefficients comes from the term in $(1-ze)^{3/2}$, leading to
$$[z^n]F(z)\sim \frac{e^{1/2}}{\sqrt{2\pi}}n^{-5/2}e^{n}.$$
That coefficient is the number of unrooted labeled trees divided by $n!$, so that multiplying by Stirling's formula gives the desired $e^{1/2}n^{n-2}$.