Sum-Product of Random Variables

Let $\Sigma_1=\sum_{i=1}^{\infty}{(\prod_{j=1}^{j}{X_i})}$ and notice that $\Sigma_1=X_1(1+\sum_{i=2}^{\infty}{(\prod_{j=2}^{j}{X_i})})=X_1(1+\Sigma_2)$

Also note that $\Sigma_1$ and $\Sigma_2$ are also iid (can be seen via dummy index changes).

Let us also assume that $\Sigma_i$ has the pdf $f_{\Sigma}(\sigma)$

Define $Y=(1+\Sigma_2)$, $f_Y(y)=f_{\Sigma}(y-1)$

Anyway since $\Sigma_2$ does not contain $X_1$, they are also independent.

Finally we can write down:

$f_{\Sigma}(\sigma)=\int_{\mathscr{Y}}{f_X(\sigma/y)f_Y(y)dy}=\int_{x=0}^{1}{f_X(x)f_Y(\sigma/x)dx}=\int_{x=0}^{1}{f_X(x)f_{\Sigma}(\sigma/x-1)dx}$

For an arbitrary $f_X(x)$ this cannot be solved, even existence of $f_{\Sigma}$ cannot be guaranteed.

One trivial solution for $(X,\Sigma)$ pair is satisfied by $f_X(t)=f_\Sigma(t)=\delta(t)$

One trivial $X$ implying that $\Sigma$ does not have a (not heavy tailed) pdf is: $f_X(t)=\delta(t-1)$

So if you can give us $f_X(x)$, we can talk again about the solution. If not for all I care I have given you a solution and an answer.