Expansion of a function with a negative fractional power How would you expand a moment generating function with a negative fractional power as a power series??
What method would you be able to use?
 A: See this, in particular equations (17)-(24). This is the moment-generating function of a gamma distribution, and so having the moments you are actually done.
EDIT:
Specifically, take $\theta = 2$ and $\alpha = n/2$. Then, the moments $\mu'_r$ are given by 
$$
\mu'_r = \frac{{2^r \Gamma (n/2 + r)}}{{\Gamma (n/2)}} = n(n+2)(n+4) \cdots (n+2r-2),
$$
where for the last equality we have used $\Gamma(p+1) = p \Gamma(p)$.
Now, the moment-generating function can be expanded as
$$
M(t) = 1 + \mu' _1 t + \frac{1}{{2!}}\mu' _2 t^2  + \frac{1}{{3!}}\mu' _3 t^3  +  \cdots,
$$
giving us
$$
\frac{1}{{(1 - 2t)^{n/2} }} = 1 + nt + \frac{{n(n + 2)}}{{2!}}t^2  + \frac{{n(n + 2)(n + 4)}}{{3!}}t^3  +  \cdots 
$$
(for $|t|<1/2$).
EDIT 2:
For completeness, let us show how to find the moments $\mu'_r$. Your moment-generating function corresponds to a gamma distribution with density
$$
f(x)= \frac{{x^{\alpha  - 1} e^{ - x/\theta } }}{{\theta ^\alpha  \Gamma (\alpha )}}, \;\; x>0,
$$
where $\theta = 2$ and $\alpha = n/2$. So,
$$
\mu'_r = \frac{1}{{\theta ^\alpha  \Gamma (\alpha )}}\int_0^\infty  {x^{r + \alpha  - 1} e^{ - x/\theta } \,{\rm d}x} = 
\frac{{\theta ^r }}{{\Gamma (\alpha )}}\int_0^\infty  {x^{r + \alpha  - 1} e^{ - x} \,{\rm d}x} = \theta ^r \frac{{\Gamma (r + \alpha )}}{{\Gamma (\alpha )}}.
$$
A: http://en.wikipedia.org/wiki/Binomial_series
A: Im trying to work out a similair question, but I am unsure as to how to expand it as a power series?
I have taken the log of my function and differentiated it, but what is my final answer for the power series expansion? 
