Suppose the moment generating function of $X$ is $$\sum_{n=1}^\infty \frac{6}{\pi^2}\frac{1}{n^2}e^{tn}\quad \text{for}\ t\le0$$ What is the distribution of $X$?
We know that the MGF for a discrete $X$ is $\sum_{x} e^{tn} p_X(x)$, where $p_X(x)$ is the PMF of $X$.
So I am assuming that the PMF of $X$ is simply $\frac{6}{\pi^2}\frac{1}{n^2}$ for all discrete $n$. What distibution does this correspond to? Or am I missing something important?
Yes, I think you are right. $p_X(n) = \frac{6}{\pi^2}\frac{1}{n^2}$ is a valid PMF for discrete variable because $\sum_{n=1}^{\infty}\frac{1}{n^2} =\frac{\pi^2}{6}$, which implies $p_X(n)$ sums up to 1 and is a valid PMF.