From Hogg, McKean & Craig 6e; Problem 3.3.4 (I'm studying for a comprehensive exam in Math Stats, so this is almost like homework; I'd prefer answers that point me towards concepts I should learn.):
"Let X be a random variable such that $E(X^m) = (m+1)!2^m$, m=1,2,3,.... Determine the mgf and distribution of X."
1) In a similar, previous example (3.3.2), they start with $E(X^m) = {(m + 3)! \over 3!} 3^m, m=1,2,3...$ and immediately state that "then the mgf of X is given by the series $$M(t) = 1 + {4!3 \over 3!1!}t + {5!3^2 \over 3!2!}t^2 + {6!3^3 \over 3!3!}t^3 + ...$$
How do they go from the expectation of $X^m$ to the mgf in either case? Is there a direct relationship between $E(X^m)$ and $E(e^{tX})$?
2) In the same example (3.3.2), once the series is created, they recognize it as the Maclaurin's series for $(1-3t)^{-4}$. How do they go from a Maclaurin's series and work back to its corresponding function?
3) I know how to get to the distribution once I know the mgf; the form of the mgf itself will indicate which distribution this random variable takes.
Yes. We have $$\mathrm{E}[X^m] = \frac{d^m}{dt^m}\left[M_X(t)\right]_{t=0},$$ from the series expansion $$M_X(t) = M_X(0) + M_X'(0) \frac{t}{1!} + M_X''(0) \frac{t^2}{2!} + M_X'''(0) \frac{t^3}{3!} + \cdots$$ and $$\begin{align*} \mathrm{E}[e^{tX}] &= \mathrm{E}\left[1 + X \frac{t}{1!} + X^2 \frac{t^2}{2!} + X^3 \frac{t^3}{3!} + \cdots \right] \\ &= 1 + \mathrm{E}[X] \frac{t}{1!} + \mathrm{E}[X^2] \frac{t^2}{2!} + \mathrm{E}[X^3] \frac{t^3}{3!} + \cdots. \end{align*}$$
This is something from undergraduate calculus; you can do it a variety of ways, but one convenient method is via differentiation of formal power series: $$\frac{1}{1-z} = (1-z)^{-1} = \sum_{k=0}^\infty z^k,$$ so by repeated differentiation, $$n! (1-z)^{-n-1} = \sum_{k=n}^\infty \frac{k!}{(k-n)!} z^{k-n}.$$