$Q)$ $E(X^n) = (-1)^n\frac{1}{5} + \frac{2}{5}$ with $n=1,2,3,....$ for discrete random variable, $X$.
Find the probability mass function for the $X$.
In the solution sheet it said $M(t) = M(0) + \sum_{r=0}^{\infty} \frac{M^{(r)}(0)}{r!} t^r = M(0) + \sum_{r=0}^{\infty} \frac{3}{5}\frac{t^{2k}}{(2k)!} + \sum_{r=0}^{\infty} \frac{1}{5}\frac{t^{2k+1}}{(2k+1)!}$
Hence, $M(t) = \frac{2}{5} +\frac{3}{5} \cosh t + \frac{1}{5} \sinh t $
The answer sheet claimed the $M(0) = \frac{2}{5}$. But Why is the $M(0)$ is $ \frac{2}{5}$? In my thought $M(0) = E(1)$ because of the $E(e^{tx}) = M(t)$.
But I can't find the $M(0)$ because the $E(X^n)$ is not defined at the $n=0$.