= Solution
For an integrable <discrete random variable>, its <expectation> and <variance> are $\mu=\sum_{n\in\mathbb Z}np_n$ and $\sum_{n\in\mathbb Z}(n-\mu)^2p_n$, respectively. The first formula requires $\sum|n|p_n<\infty$ for a finite ordinary mean; the variance may be infinite. A <probability density function> gives $P(\alpha<Y<\beta)=\int_\alpha^\beta f(y)\,dy$.
Conditioning on $X$ and using <independence> gives
$$
P(\alpha<X+Y<\beta)=\sum_{n\in\mathbb Z}p_n\int_{\alpha-n}^{\beta-n}f(y)\,dy
=\int_\alpha^\beta\sum_{n\in\mathbb Z}p_nf(z-n)\,dz.
$$
The interchange is justified by <Tonelli's theorem>, since all integrands are nonnegative. Consequently $Z$ has an absolutely continuous distribution, with
$$
\boxed{g(z)=\sum_{n\in\mathbb Z}p_nf(z-n).}
$$
Its integral is $\sum_np_n=1$. This is the <convolution of independent random variables>, specifically a <Gaussian smoothing of an integer-valued distribution> for the given normal density. The specified weights are normalized because $2(3/\pi^2)\sum_{n\geq1}n^{-2}=1$.
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