Let $\{X_n\}$ be a sequence of independet random variables with $X_n \sim U[0,1]$, $\forall n \in \mathbb{N}$ and define $$X:=\sum_{n=1}^{\infty} \prod_{i=1}^n X_i$$ Show that $\mathbb{E}[X] < \infty$
First i define $Y_n:=\prod_{i=1}^n |X_i|$ then $Y_n \geq 0, \forall n$ and $$ \mathbb{E}[Y_n]=\mathbb{E}\left[\prod_{i=1}^n |X_i| \right]=\prod_{i=1}^n \mathbb{E}[|X_i|]=\prod_{i=1}^n \frac{1}{2}=\frac{1}{2^n} $$ Where the second equality follows from the independence of the $X_i$ and the third equality follows because $$\mathbb{E}[|X_i|]=\mathbb{E}[X_i]$$ since $X_i \sim U[0,1]$ and hence
$$\mathbb{E} \left[ \sum_{n=1}^{\infty} Y_n \right]= \sum_{n=1}^{\infty} \mathbb{E}[Y_n]=\sum_{n=1}^{\infty} \frac{1}{2^n}=1$$ But from these calculations I am not sure how to conclude, any hint or help i will be very grateful.
$E(\log X_1)=-1$ therefore from the law of large numbers $$Y_n^{1/n}=(X_1\ldots X_n)^{1/n}\rightarrow e^{-1}<1$$ which implies $\sum_1^{\infty}Y_n<\infty$ almost surely from Cauchy criteria. The expectation of $\sum_1^{\infty}Y_n$ is $1$ as correctly explained in other answers.
However, the point above is necessary : if $\Pr(Y_n=1/n)=1/n$ and $\Pr(Y_n=0)=(n-1)/n$ then $\sum_{n=1}^{\infty}E(Y_n)<\infty$ while by Borel Cantelli $\sum_{n=1}^{\infty}Y_n=\infty.$