$a_1=1$ and $a_n$ is randomly $a_{n-1}$ or $\frac12 a_{n-1}$. What is $\mathbb{P}[\sum a_n<\infty]$? Define a series of random variables by $a_1=1$, and $a_n$ is either $a_{n-1}$ or $\frac12 a_{n-1}$ with equal probabilities. 

What is the probability that the series $\sum a_i$ converges?

This is a tail event, thus by kolmogorov's 0-1 law this probability is either 0 or 1.
The expectation of the sum is $$S=\sum_{n\geq 1} \mathbb{E}[a_i]=1+ \sum_{n \geq2} \mathbb{E}[\mathbb{E}[a_i|a_{i-1}]]=1+\sum_{n \geq2} \mathbb{E}[\frac34 a_{i-1}]=1+\frac34 S$$ and thus the expected value is $4$. Can we conclude from this that the series converges with probability 1?
 A: This question was solved in the comments. Here's how I interpreted it.
First, the calculation of the expectation as written does not rule out $S=\infty$, and should be fixed by noting by induction that $\mathbb{E}[a_i]=(\frac34)^{i-1}$, since $\mathbb{E}[a_i] = \frac12 \mathbb{E}[a_{i-1}]+ \frac12 \mathbb{E}[\frac12a_{i-1}]=\frac34 \mathbb{E}[a_{i-1}]$. Then by a geometric series we get $\mathbb{E}[S]=4$.
Second, Since the sum $S$ is a non-negative random variable, we can invoke markov's inequality: We truncate $S$ to have value $5$ whenever it diverges to make all its values finite. This can only lower the expectation of $S$. (We truncate to avoid justifying the use of markov with a r.v. that does not get values in $\mathbb{R}$.) 
So we have $4\geq\mathbb{E}[S]\geq 4.5\times\mathbb{P}[S\geq 4.5] $, and thus in particular the probabilty that sum diverges before our alteration is smaller than $1$, and by kolmogorov it is $0$.
A similar argument works whenever $\mathbb{E}[|X|]$ is finite:
We look at $|X|$, truncate it to  $n\mathbb{E}[|X|]$ when it diverges, and by Markov's inequality we get
 we have $\mathbb{P}[|X| \text{ diverges}] \leq \mathbb{P}[|X| \geq n\mathbb{E}[|X|]] \leq \frac1n$ for all $n$, hence it is $0$, and $X$ is finite a.e.
A: Consider the finite series $\sum_n^M a_n$ for $M > 1$ which has a maximum of $M$ 
max$\sum_n^M a_n = M$ for  $M > 1$ 
$\mathbb{P}\big[\sum_n^M a_n=M\big]=\big(\frac{1}{2} \big)^{M-1}$ 
$\lim_{M\to \infty}\mathbb{P}\big[\sum_n^M a_n=M\big]=\lim_{M\to \infty}(1/2)^{M-1}=0$ 
$\Longrightarrow \mathbb{P}\big(\sum_n^M a_n<\infty\big) = 1-\lim_{M\to \infty}\mathbb{P}\big[\sum_n^M a_n=M\big]=1-0=1$
