How to calculate the geometric moments of a log-normal distribution? If I only know geometric mean and geometric standard deviation of a log-normal distribution how can I calculate the $n$-th moment of the distribution?
In the Wikipedia article I can only see a relationship for the $n$-th moment if the (arithmetic) mean and standard deviation are known:
$$
E\left[X^n\right]=e^{n\mu+\frac{1}{2}n^2\sigma^2}
$$
If only the geometric mean and geometric standard deviation are known, is there still a way to calculate the moments?
The geometric mean and standard deviation are defined by:
$$
GM = e^{\mu_l}\quad\mathrm{where}\quad\mu_l = \frac{\sum_{i=1}^N\ln(x_i)}{N} 
$$
and
$$
GSD = e^{\sigma_l}\quad\mathrm{where}\quad\sigma_l=\sqrt{\frac{\sum_{i=1}^N\left[\ln(x_i)-\mu_l\right]^2}{N}}
$$

For simplicity I will call $\mathcal{N}_k$ the $k$-th moment of the normal distribution with parameters $\mu$ and $\sigma$, so for example $\mathcal{N}_1 = \mu$, $\mathcal{N}_2 = \mu^2 + \sigma^2$, $\cdots$
Now if $X$ follows a lognormal distribution, then
\begin{eqnarray}
\mathbb{E}[\ln^k X] &=& \int\frac{{\rm d}x~}{x}\frac{\ln^k x}{\sigma\sqrt{2\pi}} \exp\left[-\left(\frac{\ln x - \mu}{2\sigma^2}\right)^2\right] \\
&\stackrel{y=\ln x}{=}&\int{\rm d}y ~\frac{y^k}{\sigma\sqrt{2\pi}}\exp\left[-\left(\frac{y-\mu}{2\sigma^2}\right)^2 \right] \\
&=& \mathcal{N}_k \tag{1}
\end{eqnarray}
In you case, you have then
$$
\mu_l = \ln{\rm GM} = \mathbb{E}[\ln X] = \mathcal{N}_1 = \mu
$$
and
$$
\sigma_l^2 = \mathbb{E}[(\ln X - \mu_l)^2] = (\cdots) = \sigma^2
$$
$\mu_l$ and $\sigma_l$ thus allow you to estimate $\mu$ and $\sigma$
