moments of gaussian distribution by taylor series expansion I have calculated characteristic function of the normal distribution 
$$f_X(k)=\exp\left(ika-\frac{\sigma ^{2}k^{2}}{2}\right)$$ and now I would like to find the moments, so I know that you could expand characteristic function by Taylor series

$$
\begin{split}
f_X(k)
  &amp;= \left(1 + \frac{1}{1!}\left(ika - \frac{\sigma^2k^2}{2}\right)
                 + \frac{1}{2!}\left(ika - \frac{\sigma^2k^2}{2}\right)^2
                 + \frac{1}{3!}\left(ika - \frac{\sigma^2k^2}{2}\right)^3
                 + \ldots \right)\\
  &amp;= \left(1+\frac{(ik)}{1!}\left \langle X^1 \right \rangle+\frac{(ik)^2}{2!}\left \langle X^2 \right \rangle+\frac{(ik)^3}{3!}\left \langle X^3 \right \rangle+...\right)
\end{split}
$$

and the moments will be
$\left \langle X^n \right \rangle$

Now the problem is that I completely forgot how to evaluate Taylor series.
Could you be so kind and help me to calculate for example second moment? I know what the answer should be, but I couldn't get it right.

Any help would be appreciated!
I'm not sure we need to use Taylor series to compute moments given the moment generating function.

Let $X$ be RV with MGF:

$$M_X(t) := E[e^{Xt}]$$

We have

$$M_X^{'}(0) = E[Xe^{X(0)}], M_X^{''}(0) = E[X^2e^{X(0)}]$$



Now if you wanted to get the mgf given the moments $E[X^n]$ do this:

$$E[e^{Xt}] = E[\sum_{n=0}^{\infty} \frac{X^n}{n!}]$$
$$= \sum_{n=0}^{\infty} E[\frac{X^n}{n!}]$$
$$ = \sum_{n=0}^{\infty} \frac{1}{n!} E[X^n]$$
I'm not sure we need to use Taylor series to compute moments given the moment generating function.

Let $X$ be RV with MGF:

$$M_X(t) := E[e^{Xt}]$$

We have

$$M_X^{'}(0) = E[Xe^{X(0)}], M_X^{''}(0) = E[X^2e^{X(0)}]$$



Now if you wanted to get the mgf given the moments $E[X^n]$ do this:

$$E[e^{Xt}] = E[\sum_{n=0}^{\infty} \frac{X^n}{n!}]$$
$$= \sum_{n=0}^{\infty} E[\frac{X^n}{n!}]$$
$$ = \sum_{n=0}^{\infty} \frac{1}{n!} E[X^n]$$