In the Wikipedia article on the Weibull distribution, the moment generating function is given:
$$g(s) = \sum\limits_{n=0}^\infty \frac{s^n \lambda^n}{n!} \Gamma\left(1+\frac{n}{\kappa}\right )$$
However, it is only defined for $\kappa \geq 1$. Doesn't seem like anything will break here for $\kappa < 1$. And that is an interesting case where it has a decreasing hazard rate. So, why can't I apply this formula for $\kappa < 1$?
So the Weibull pdf is defined as $$f:\mathbb R_+\to \mathbb R_+~~,~~f(x;\lambda,\kappa)=\frac{\kappa}{\lambda}(x/\lambda)^{\kappa-1}\exp\big((-x/\lambda)^\kappa\big)$$ WLOG let's consider $\lambda=1$: $$\phi(x;k):=f(x,1,k)=k x^{k-1}\exp(-x^k)$$ The moment generating function is defined as a sort of Laplace transform: $$g(s;k):=(\mathcal Lf)(-s;k)=\int_0^\infty e^{sx}\phi(x;k)\mathrm dx$$ This integral is $$k\int_0^\infty x^{k-1}\exp(sx-x^k)\mathrm dx$$ But what happens if $k<1$? Well, then actually the integral fails to converge, since the argument of the exponential function $sx-x^k$ will grow without bound, since $x^1$ grows faster than $x^{1/2}$, e.g. So our integrand looks like $$\frac{ \exp(O(x^1))}{x^{1-k}}$$ And of course the exponential growth in the numerator beats the power growth in the denominator. As for the sum representation, it is simply a Taylor series. By Taylor's theorem we know that in some neighborhood of $s=0$ that $$g(s;k)=\sum_{n=0}^\infty\frac{(\mathrm D^{n}g)(0;k)}{n!}x^n$$ And one can see from the Leibniz integral rule that $$(\mathrm D^ng)(0;k)=k\int_0^\infty x^{k+n-1}\exp(-x^k)\mathrm dx$$ With a change of variable (i.e $z=x^k$) one can see that this is just $$(\mathrm D^ng)(0;k)=k\int_0^\infty x^{k+n-1}\exp(-x^k)\mathrm dx=\Gamma(1+n/k)$$ Which gives us $$g(s;k)=\sum_{n=0}^\infty\frac{s^n}{n!}\Gamma(1+n/k)$$