= Solution
Choose a positive integer-valued <random variable> with <probability mass function>
$$
\boxed{\mathbb P(X=k)=\frac4{k(k+1)(k+2)},\qquad k=1,2,\ldots.}
$$
These masses are positive. They sum to one because the <partial fraction decomposition>
$$
\frac4{k(k+1)(k+2)}=2\left(\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right)
$$
makes their sum a <telescoping series> with value $2/(1\cdot2)=1$. Its <expectation> is another <telescoping series>:
$$
\mathbb E X=4\sum_{k=1}^\infty\frac1{(k+1)(k+2)}
=4\sum_{k=1}^\infty\left(\frac1{k+1}-\frac1{k+2}\right)=2.
$$
By contrast, since $k+1\leq2k$ and $k+2\leq3k$ for $k\geq1$, its <second moment> satisfies
$$
\mathbb E X^2=4\sum_{k=1}^\infty\frac{k}{(k+1)(k+2)}
\geq\frac23\sum_{k=1}^\infty\frac1k=\infty.
$$
The comparison uses the divergent <harmonic series>. To check the <variance> directly, for $k\geq4$ we have $(k-2)^2\geq k^2/4$, so the corresponding tail of $\mathbb E[(X-2)^2]$ also diverges. Thus
$$
\boxed{\mathbb E X=2<\infty,\qquad\operatorname{Var}(X)=\infty.}
$$
Here “unbounded” means infinite <variance> for this single <probability distribution>.
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