Assume that $(\gamma_i)$ is a square summable sequence of real numbers and let $(W_i)$ be an i.i.d. sequence standard normal distributed random variables defined on $(\Omega,\mathbb{F},P)$.
One can show that the sequence $(L_n)$ defined by $$ L_n =\sum_{i=1}^n \gamma_i (W_i^2-1) \quad \quad \forall n\in \mathbb{N} $$ is Cauchy in $L^2(P)$ and therefore converges to some $Z\in L^2(P)$.
How do i show that the limit $Z$ has a continuous cdf?
If $X,Y$ are RVs and distributed independently, then the CDF of $X+Y$ may be denoted as following, $$F_{X+Y}(z)=\int_{-\infty}^{+\infty}F_X(z-y)\mathrm{d}F_Y(y).$$ Hence $X+Y$ has continuous (or absolutely continous) CDF, if $X$ has continuous (or absolutely continous resp.) CDF. Now $$ Z=\sum_{i=1}^\infty\gamma_i(W^2_i-1)=\gamma_1(W^2_1-1)+\sum_{i=2}^\infty\gamma_i(W^2_i-1),$$ meanwhile the $W_1^2$ is $\chi^2(1)$-distributed and has (absolutely) continuous CDF, also $\gamma_1(W^2_1-1)$ is independent with others, therefore $Z$ has a (absolute) continuous CDF.