If $(1-\frac x 1 + \frac {x^2} 2 -\cdots)^{-1} = A_0 + \frac{A_1x^2} {1!} + \frac {A_2x^2}{2!}+\cdots$ then $A_n \sim (-1)^{n-1}(n-1)!(\log n)^{-2}$ From G.Pólya "Mathematics and Plausible Reasoning" p.9. Problem 8:
Set $$\biggl(1-\frac x 1 + \frac {x^2} 2 -\frac {x^3} 3 +\cdots \biggr)^{-1} = A_0 + \frac{A_1x^2} {1!} + \frac {A_2x^2}{2!}+\cdots$$
We find for $$n = 0 \phantom{2}1 \phantom{2}2 \phantom{2}3\phantom{2}4\phantom{32}5\phantom{42}6\phantom{432}7\phantom{452}8\phantom{452}9$$ $$A_n = 1\phantom{2}1\phantom{2}1\phantom{2} 2\phantom{2}4\phantom{2}14\phantom{2}38\phantom{2}216\phantom{2}600\phantom{2}6240.$$
Then the answer says: "By more advanced tools (integral calculus, or theory of analytic functions 
of a complex variable) we can prove that, for large $n$, the value of $A_n$ is approximately $(-1)^{n-1}(n-1)!(\log n)^{-2}$."
How is this approximation obtained?  I tried complex analysis. The left part of the equation equals $$\frac 1 {1-\log(1+z)} (\lvert z \rvert<1).$$
From the answer I guess there isn't a closed form of $A_n$.  
 A: $$ f(z)=\frac{1}{1-\log(1+z)} $$
is an analytic function in a neighbourhood of the origin and the radius of convergence of the Taylor series at the origin equals one. By considering 
$$ f(z) = \sum_{n\geq 0}a_n z^n = 1+z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{6}+\frac{7 z^5}{60}+\ldots$$
$$ f'(z) = \frac{f(z)^2}{z+1}\quad\text{(+ repeated differentiation)} $$
we get $a_n>0$ for any $n\in[0,11]$. By Cauchy's integral formula
$$ a_n = \frac{1}{2\pi i}\oint_{|z|=\varepsilon}\frac{f(z)}{z^{n+1}}\,dz =\frac{1}{2\pi i}\oint_{|z|=\varepsilon}\frac{f(e^t-1)}{(e^t-1)^{n+1}}e^t\,dt=\frac{1}{2n\pi i}\oint_{|z|=\varepsilon}\frac{dt}{(1-t)^2(e^t-1)^{n}}$$
hence for any $n\geq 1$
$$ a_n-a_{n-1} = \text{Res}\left(\frac{1}{(1-t)(e^t-1)^n},t=0\right) $$
$$ a_n= \frac{1}{n}\cdot\text{Res}\left(\frac{1}{(1-t)^2(e^t-1)^n},t=0\right) $$
and the magnitude of $a_n$ can be estimated through Laplace method. A similar behaviour is shown by the coefficients of the Taylor series at the origin of $\frac{z}{\log(1+z)}$, also known as Gregory coefficients.
About them, this article from Blaglouchine is enlightening.
