I tried this way: As the sum of convergent geometric series is $\frac{a}{1-r}$ and $-1<r<1$.
Moreover sum is also a rational number. So $a$ and $r$ should be rational numbers. As rational numbers are countable. We can say $a$ and $r$ are countable, which means the set of series is also countable.
But my doubt is if $a= 0.333.....$ and $r= 0.7777....$ , where both are irrationals. Sum will be rational in that case also $(0.333/1-0.777=1)$. In that case $a$ and $r$ need not be rational which means they will be uncountable.
I am stuck here.
I believe the set is uncountable. I will make use of the fact that the set of irrational numbers $x$ such that $0<x<1$ is uncountable.
Let $r$ be an irrational number between $0$ and $1$. $r$ will be the common ratio of the series.
Now, we want to find a number $u$ to be the first term of the series, such that the sum of the series is rational.
That is, we need $\frac{u}{1-r}=\frac{p}{q}$ for some integers $p$ and $q$.
Well, just let $u=\frac{p}{q}\times(1-r)$. Then the sum of the series will be $\frac{p}{q}$. Note that $u$ is also an irrational number, since $r$ is irrational.
Since there are an uncountable number of choices for $r$, there are an uncountable number of geometric series with rational sum.