# Are the set of all convergent geometric series whose sum is a rational number is countable? [closed]

I tried this way: As the sum of convergent geometric series is $$\frac{a}{1-r}$$ and $$-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.

• Here is a quick guide to MathJax to format your posts on this website : math.meta.stackexchange.com/questions/5020/… – Saket Gurjar May 28 at 19:19
• I think you didn't finish writing your question. – Noah Schweber May 28 at 19:32
• Are the set of all convergent geometric series whose sum is a rational number is countable or not? That's my question sir – Sohith Deva May 28 at 19:37
• @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. – KReiser May 29 at 1:10
• @SaketGurjar huh, that's weird. Thanks for the explanation, and my apologies if I came off too accusative. – KReiser May 29 at 16:06

I believe the set is uncountable. I will make use of the fact that the set of irrational numbers $$x$$ such that $$0 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.

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 $$-1, $$(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.