Series of continuous random variables is continuous We work on the usual $(\Omega,\mathscr{F},P)$. Suppose $X_i$ are independent random variables. Say the distribution of $X_i$ is $F_i$. Under what circumstances can I guarantee that $\sum_{i=1}^\infty X_i$ is a continuous random variable?
Is assuming each $F_i$ is continuous enough? Is absolutely continuous enough? Is assuming $\sum_{i=1}^\infty X_i$ exists a.s. enough? What about all $X_i$ nonnegative or in $L^1$? What about if the sum in in $L^1$?
 A: Here I will say that $X$ is continuous or absolutely continuous if its distribution function $F$ has the corresponding property.  $X$ is continuous iff $P(X=x) =0$ for every $x$; it is absolutely continuous iff $P(X \in E) =0$ for every Lebesgue-null set $E$.
Fact. Suppose $X,Y$ are independent random variables.  If $X$ is continuous, so is $X+Y$.  If $X$ is absolutely continuous, so is $X+Y$.  
No assumptions about the distribution of $Y$ are needed!
Thus, in your setup, if the sum $\sum X_i$ converges (a.s., i.p. or in $L^p$), and even a single one of the $X_i$ is continuous (respectively, absolutely continuous), then the same is true of the sum.  (Just take $Y$ to be the sum of all the remaining $X_i$.)
The proof of the fact is just the convolution formula: if $X,Y$ are independent, and the distribution function of $Y$ is $G$, then $P(X+Y \in A) = \int P(X \in A-y)\, G(dy)$.  If $X$ is continuous and $A = \{x\}$, then $A-y = \{x-y\}$ and so $P(X \in A-y) = P(X = x-y) = 0$ for every $y$.  Thus the integral vanishes.  Similarly, if $X$ is absolutely continuous and $A$ is Lebesgue null, then $A-y$ is also Lebesgue null for every $y$, and so $P(X \in A-y)=0$ for every $y$.
This is an illustration of the general principle that the convolution of two things is at least as smooth as whichever of the two things is smoother.  For instance, if $X$ has a density that is $k$ times continuously differentiable, then so does $X+Y$.
For the questions in your last few sentences: think about an example where $X_1$ has a discrete, nonnegative, $L^1$ distribution (perhaps a coin flip), and $X_2 = X_3 = \dots = 0$.  You aren't going to get continuity of the sum solely from the integrability of the summands, nor from the mode in which the sum converges.  There has to be some continuity in one of the summands for you to work with.
