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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.

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  • $\begingroup$ Here is a quick guide to MathJax to format your posts on this website : math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Saket Gurjar May 28 at 19:19
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    $\begingroup$ I think you didn't finish writing your question. $\endgroup$ – Noah Schweber May 28 at 19:32
  • $\begingroup$ Are the set of all convergent geometric series whose sum is a rational number is countable or not? That's my question sir $\endgroup$ – Sohith Deva May 28 at 19:37
  • $\begingroup$ @SaketGurjar the edit history makes it look like you put in almost all of the relevant mathematical content of this question. If true, I think this is probably inappropriate - please consider whether you are really preserving the goals of the post's owner. Also, the mathematical content of the edit is wrong - both those numbers are rational and they don't sum to 1. $\endgroup$ – KReiser May 29 at 1:10
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    $\begingroup$ @SaketGurjar huh, that's weird. Thanks for the explanation, and my apologies if I came off too accusative. $\endgroup$ – KReiser May 29 at 16:06
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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.

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DreiCleaner's answer is valid, in the somewhat stilted sense that if $R$ is rational and $a$ is irrational, then $a R$ is irrational. I don't think multiplication by $a$ should matter in calling something a geometric series. Therefore, if $r$ is rational and $-1<r<1$, $(1-r)^{-1}$ is rational as well. In fact, this map is a bijection: $(-1,1)\cap \mathbb{Q}\to (1/2,\infty)\cap \mathbb{Q}$, which is clearly countable.

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