Try it without integers. It doesn’t work.
Let’s take an example. Let’s say n = 2.5.
First, we’d have to define how one even does sums with non-integers, how to increment the value of i. Obviously you start with the initial value (1 in the cases we’ve been dealing with), but what’s the next number? Do you go:
1, 2, 2.5?
Do you go:
1.0, 1.5, 2.0, 2.5?
Do you go:
1.0, 1.1, 1.2 … 2.4, 2.5?
Do you stop at 2 (because you can’t jump by a whole integer amount from 2 to 2.5)? Do you stop at 3 (because maybe n = 2.5 means don’t stop as long as you’re under 2.5)?
But for argument’s sake, let’s say you go: 1, 2, 2.5. Then the sum is:
(\frac{9}{10^{1}} + \frac{9}{10^2} + \frac{9}{10^{2.5}} = \frac{9}{10} + \frac{9}{100} + \frac{9}{316.227766} = 0.9 + 0.09 + 0.02846 = 1.01846)
I’m not just saying (\infty) does not refer to a specific number, I’m saying it doesn’t refer to any number. It can’t. You’ve heard me say it before and I’ll say it again. Infinity means endless. It refers to a property of sets. It’s not that we’re not being specific about which number it refers to, it’s that we (or I) are not referring to something that could be a number.
Like I said, (\sum_{i=1}^{\infty} \frac{9}{10^i}) is not a special case of (\sum_{i=1}^{n} \frac{9}{10^i}). It’s a case that lies outside all specific cases of (\sum_{i=1}^{n} \frac{9}{10^i}) (because (\infty) is not a valid value for n). (\sum_{i=1}^{\infty} \frac{9}{10^i}) is saying: don’t pick a value for n. Let the sum run forever.
So I agree that there is no n such that (\sum_{i=1}^{n} \frac{9}{10^i} \geq 1), but (\infty) is not a valid option for n. Plugging (\infty) into the upper bound of the sum isn’t picking a value for n. It’s refusing to pick a value for n. Only in that case does the sum equal 1.
You’re so close to biting, Magnus. Wanna talk about hyperreals?