$Y$ is a random variable with $M(t) = \frac{1}{(2-\exp(t))^s}.$
Does $\frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$ converge in distribution as $s$ tends to infinity?
I let $Z = \frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$. Differentiating the MGF of $Y$ and let $t = 0$ we have $E(Y) = s,$$E(Y^2) = s^2 +2s,$$\operatorname{Var}(Y) = 2s.$ Thus Mgf of Z: \begin{align*}E(\exp(tz)) &= E\left(\exp\left(\frac{t(y-s)}{\sqrt{2s}}\right)\right)\\ &=E\left(\exp\left(\frac{ty}{\sqrt{2s}}\right)\exp\left(\frac{-ts}{\sqrt{2s}}\right)\right)\\ &=E\left(\exp\left(\frac{ty}{\sqrt{2s}}\right)\exp\left(\frac{-ts}{\sqrt{2s}}\right)\right)\\ &=\exp\left(\frac{-ts}{\sqrt{2s}}\right)\frac{1}{\left(2-\exp\left(\frac{t}{\sqrt{2s}}\right)\right)^s}. \end{align*} But I'm not sure how to proceed - seems to me it tends to $0$?