Magnus Anderson wrote:You'd have to explain why you're limiting yourself to integers.

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:

\(\frac{9}{10^{1}} + \frac{9}{10^2} + \frac{9}{10^{2.5}} = \frac{9}{10} + \frac{9}{100} + \frac{9}{316.227766} = 0.9 + 0.09 + 0.02846 = 1.01846\)

Magnus Anderson wrote:Basically, what you're saying is that \(\infty\) does not refer to a specific number in the same way that the statement "An integer greater than 10" does not refer to a specific integer since the quantity that it represents can be 11 or 12 or 100. However, you have yet to explain why it's invalid for n (the upper bound of summation) to be a non-specific number.

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.

Magnus Anderson wrote:My claim is that \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for any \(n\) that is greater than \(0\). It does not have to be an integer. The only condition is that it's greater than \(0\). Do you think there is \(n\) greater than \(0\) such that the sum is not less than \(1\)? No. You obviously do (otherwise, you'd not say that \(0.\dot9 = 1\).)

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.

Magnus Anderson wrote:\(\infty\) is not an integer. It's a number greater than every integer.

And \(\sum_{i=1}^{\infty} \frac{9}{10^i}\) is not a completely different statement. The only difference between \(\infty\) and integers is 1) \(\infty\) is a bigger number and 2) it's non-specific. The fact that it's non-specific shouldn't be a problem at all. And if you absolutely hate non-specific numbers, you can switch to some specific infinity.

You're so close to biting, Magnus. Wanna talk about hyperreals?