The moment generating function of $X$ is given by $M_X(t) = E[e^{tX}]$.
I'm wondering if it is always possible to obtain moments as $E[X^k] = M_X^{(k)}(0)$, i.e. the $k^\text{th}$ derivative of the moment evaluated at $0$. In particular, does this hold even if $X$ can take negative as well as positive values?
Say $X$ has pdf $f_X$. Using the series expansion of $e$ we have, $M_X(t) = E[e^{tX}] = E[\sum_{i=0}^{\infty}\frac{(tX)^i}{i!}] = \int_{-\infty}^{\infty} \sum_{i=0}^{\infty}\frac{(tX)^i}{i!} f_X(x) dx$. This is equal to $\int_{-\infty}^{\infty} \lim_{n\rightarrow \infty} \sum_{i=0}^{n}\frac{(tX)^i}{i!} f_X(x) dx$. If $X$ is always non-negative, then the dominated convergence theorem allows for the exchange of the order of the limit and integral. So we $E[e^{tX}] =\lim_{n\rightarrow \infty} E[\sum_{i=0}^{n}\frac{(tX)^i}{i!}] =\sum_{i=0}^{\infty} \frac{t^i}{i!} E[X^i]$. However, if $X$ is allowed to take values in $(-\infty,\infty)$, I can't see how the order of the limit and integral can be exchanged.
Thanks!
That said, the mgf $M(t)$ is the expectation of a non negative random variable $e^{tX}$; therefore it will be always well defined; however, for some t, the expectation might diverge to $+\infty$.
Clearly, $M(0)=1$. Furthermore, $M$ has the following property: if for $t_0>0$, $M(t_0)<+\infty$, then for all positive $t<t_0$, $M(t)<+\infty$. This is because $e^{tX} \leq 1 + e^{t_0X}$. The same applies if $t_0$ is negative: if the mgf is finite for a particular value $t_0$, then it is finite for all $t$ closer to $0$.
So we have two cases: either the mgf is finite in a neighbourhood of zero (i.e. exists $\delta >0$ such that $M(t)<+\infty$ for all $t \in(-\delta,\delta)$) or the mgf is infinity on one or both sides of zero.
CASE 1: If, for some $\delta>0$, the mgf is finite on $(-\delta,\delta)$, then for any $t \in (-\delta,\delta)$, $M(t)= \sum_{k=0}^{\infty} \frac{t^k E[X^k]}{k!}$.
Proof: $\sum_{k=0}^{\infty} \frac{|tX|^k}{k!} = e^{|t X|} \leq e^{t X} + e^{-t X}$. Taking the expectation of the last inequality, we have that right hand side is $M(-t)+M(t)<+\infty$ by assumption. This tells you that $E[|X|^k]<\infty$. Furthermore, noting that $|\sum_{k=0}^{n} \frac{(tX)^k}{k!} | \leq \sum_{k=0}^{n} \frac{|tX|^k}{k!} \leq \sum_{k=0}^{\infty} \frac{|tX|^k}{k!} $ and the last is integrable (we saw that above), we can use dominated convergence theorem obtaining $M(t)= \sum_{k=0}^{\infty} \frac{t^k E[X^k]}{k!} $.
Remark: Differentiating the power series term-by-term yields $E[X^k]=M^{(k)}(0)$, which is what you wanted.
CASE 2: If the mgf is infinity on one or both sides of zero, then clearly you don't have the result regarding the derivatives, as they can't exist.