If $\mathbb EX=0$ and $\mathbb E|X| < \infty $ implies finite Variance?

Question

I think, I am totally wrong, but if $\mathbb E|X|$ is finite then,

$$\text{Var}|X|=\mathbb E|X^{2}|- (\mathbb E(X))^{2}=\mathbb E|X^{2}|< \infty$$ Considering $\mathbb E|X|< \infty$ implies $\mathbb E|X^{2}|<\infty$.

Answer 1

More generally, if the $k$th moment is finite, the moment of order $k+1$ need not be finite. For instance, the $t$ distribution with $k>1$ degrees of freedom has moments of order $1,2,...,k-1$ while moments of order $k$ and above do not exist.