Let me try.

When the pdf \( f_X \) is symmetric about 0, we have

\( f_X(x) = f_X(-x) ~~ \forall x \in \mathbb{R} \)

Next we consider the odd moments

\( m_{2k-1} = E[X^{2k-1}] \)

\( = \int_{-\infty}^{+\infty} x^{2k-1}f_X(x)dx \)

\( = \int_{-\infty}^0 x^{2k-1}f_X(x)dx + \int_0^{+\infty} x^{2k-1}f_X(x)dx

\)

Consider the former integral:

\( \int_{-\infty}^0 x^{2k-1}f_X(x)dx \)

\( = \int_{-\infty}^0 x^{2k-1}f_X(-x)dx \)

Next let \( y = -x \); then when \( x \to -\infty, y \to +\infty \); \( x = 0, y = 0 \); \( dx = -dy \)

\( = \int_0^{+\infty} (-y)^{2k-1}f_X(y)dy \)

\( = -\int_0^{+\infty} y^{2k-1}f_X(y)dy \)

which cancel with the latter integral. Essentially, \( f_X(x) \) is an even function and \( x^{2k-1} \) is an odd function, therefore their product, the integrand \( x^{2k-1}f_X(x) \) is also an odd function which should integrate to 0 as expected.

The moment-generating function in this case can be expressed as

\( E[e^{tX}] = 1 + \sum_{k=1}^{+\infty} m_{2k} \frac {t^{2k}} {(2k)!} \)

since all the odd moments vanish.

Denote \( h(t) = \sum_{k=1}^{+\infty} m_{2k} \frac {t^{2k}} {(2k)!} \)

such that \( E[e^{tX}] = 1 + h(t) \)

http://en.wikipedia.org/wiki/Mercator_series
From the above result we know that the cumulant generating function is

\( g(t) = \ln E[e^{tX}] = \ln(1 + h(t))

= \sum_{n=1}^{+\infty} \frac {(-1)^{n+1}} {n}h(t)^n \)

As we know all the coefficients of odd powered terms in \( h(t) \) are 0, it should be the same for \( h(t)^n \) and thus \( g(t) \);

and by definition the cumulants \( \kappa_n \) satisfies

\( g(t) = \sum_{n=1}^{+\infty} \kappa_n \frac {t^n} {n!} \)

By comparing the coefficients, we can immediately claim that all the odd cumulants \( \kappa_{2k-1} \) vanish.

There maybe some shorter and smarter ways to show this without going through the technical details in the above arguments. (There are always some technical conditions need to be considered whenever you are dealing with sum to infinity)