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I have some trouble with this computation, I have the moment generating function of a random variable $S$ by:

$$M_S(t)=\frac{\beta\mu t}{1+(1+\beta)\mu t-M_X(t)}$$ According to the text that I am following, I need to consider the saddlepoint approximation i.e. I need to find$\theta=\theta(x)$ such $k_S(\theta)=x$, where $k_X(t)$ is the cummulant generating function, that is:$$k_S(\theta)=ln(M_S(\theta))$$.

Now, Considering these facts I have that:

$$k_S(\theta)=ln\frac{\beta\mu\theta}{1+(1+\beta)\mu\theta-M_X(\theta)}$$ in this case $X$ is a random variable with Weibull distribution. I need to compute $k_s'(\theta)$. (I have trouble with it¡¡)

$$k_S'(\theta)=\frac{1+(1+\beta)\mu\theta-M_X(\theta)}{\beta\mu\theta}\frac{\beta\mu(1+(1+\beta)\mu\theta-M_X(\theta))-((1+\beta)\mu-M_X'(\theta))\beta\mu\theta}{(1+(1+\beta)\mu\theta-M_X(\theta))^2}$$

Here, I only use the derivative properties for quotient and the derivative of the logarithm function.

Then, by symplifying I got: $$k_S'(\theta)=\frac{1}{\beta\mu\theta} \frac{\beta\mu(1+\mu\theta+\beta\mu\theta-M_X(\theta))-(\mu+\beta\mu-M_X'(\theta))\beta\mu\theta}{(1+(1+\beta)\mu\theta-M_X'(\theta))}$$ $$k_S'(\theta)=\frac{\beta\mu+\beta\mu^2\theta+\beta^2\mu^2\theta-\beta\mu M_X(\theta)-\beta\mu^2\theta-\beta^2\mu^2\theta+\beta\mu\theta M _X(\theta)}{\beta\mu\theta(1+(1+\beta)\mu\theta-M_X(\theta))}$$

$$k_S'(\theta)=\frac{\beta\mu-\beta\mu M_X(\theta)+\beta\mu\theta M_X'(\theta)}{\beta\mu\theta((1+(1+\beta)\mu\theta-M_X(\theta))}$$ $$k_S'(\theta)=\frac{1-M_X(\theta)+\theta M_X'(\theta)}{1+(1+\beta)\mu\theta-M_X(\theta)}$$ Now, since the m.g.f of a Weibull distribution is: $$M_X(\theta)=\sum_{n=0}^{\infty}\frac{\lambda^n\theta^n}{n!}\Gamma(1+\frac{n}{k})$$ and

$$\theta M'_X(\theta)=\theta\sum_{n=0}^{\infty}\frac{\lambda^nn\theta^{n-1}}{n(n-1)!}\Gamma(1+\frac{n}{k})=\sum_{n=1}^{\infty}\frac{\lambda^n\theta^n}{(n-1)!}\Gamma(1+\frac{n}{k})$$

However, I can´t conclude my computations in order to get: $\theta=\theta(x)$ such that $k_S'(\theta)=x$

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