Are we guaranteed that the harmonic series minus infinite random terms always converge? Consider the known harmonic series
$\sum_{n=1}^\infty \frac{1}{n}$ and modify it as follows
$$\sum_{n=1}^\infty a_n\frac{1}{n}$$
where
$$a_n \sim \operatorname{Bern} \left({\frac{1}{2}}\right)$$
i.e. each $a_n$ is $0$ or $1$ with probability $\frac{1}{2}$ (basically what one is doing here is removing randomly an infinite number of terms of this series)
Is it possible to know that this series is almost always convergent? Meaning that it converges with probability $1$?

This answer combines and elaborates on two comments to the OP's post. Let 
$$a_n \sim \operatorname{Bern} \left({f(n)}\right),\;\;\; f(n) \in (0,1)$$
set $X_n \equiv n^{-1}a_n$ and consider 
$$\sum_{i=1}^\infty X_i  $$
with the random variables assumed independent.
Checking the conditions of Kolmogorov's Three-series Theorem, set $A > 1$. Then 
$$\Pr(|X_n|\geq A)=0 \implies \sum_{i=1}^{\infty} \Pr(|X_i|\geq A) =0 \tag{1}$$
converges (this is just a special case that holds in general for any random variable with support bounded from above). 
Also $I_{\{|X_n|\leq A\}} =1\;\; \forall\, n$, so
$$\sum_{i=1}^{\infty}E[X_i\cdot I_{\{|X_i|\leq A\}}] = \sum_{i=1}^{\infty}E[X_i] =\sum_{i=1}^{\infty}\frac {f(i)}{i}  \tag{2}$$
A choice for $f(n)$ for this to converge is $f(n) = n^{-\alpha},\, \alpha >0$, since this will give a hyperharmoninc series, which converges.  
Finally, 
$$\sum_{i=1}^{\infty}\text{Var}[X_i\cdot I_{\{|X_i|\leq A\}}] = \sum_{i=1}^{\infty}\text{Var}[X_i] =\sum_{i=1}^{\infty}\frac {f(i)[1-f(i)]}{i^2} \tag {3}$$
which converges for all and any $f(i) \in (0,1)$, since it is strictly smaller than $\sum_{i=1}^{\infty}i^{-2} = \pi^2/6$.
So the critical condition, is the convergence of the sum of the expected values (condition $2$), since it is this condition that imposes restrictions on the form of $f(n)$ in order to obtain convergence.
