Suppose I have a sequence of positive continuous random variables $\{X_k\}_{k=1}^\infty$ with MGF's $M_{X_k}(s)$. Furthermore, it is known that \begin{equation}\frac{X_n-n\mu}{\sqrt{n}\sigma}\rightarrow\mathcal{N}(0,1),\end{equation} for some known $\mu$ and unknown $\sigma$. Given the function \begin{equation}F[z,s]=\sum_{n=0}^\infty z^{-n} M_{X_n}(s),\end{equation} is it possible to extract $\sigma$ without the use of inverse transforms?
For example: \begin{equation}F[z,s]=\frac{zs}{1-e^s+zs}.\end{equation} Answer: $\sigma^2=\frac{1}{12}$.
I think your example is wrong, because moment generating functions should be 1 when s is 0, but when s is 0, your example is 0.
In general, if you take the second derivative with respect to s and set s=0, you should get the z-transform of the variance; call it g. To find its limit, use the final limit theorem (http://en.m.wikipedia.org/wiki/Z-transform#Properties) by taking $\lim_{z\rightarrow 1}(z-1)g$.