S2 is identical to (0 + S1). Are you suggesting that (0 + S1) != S1 ?

Precisely. Adding 0 to the front of an infinite series is shifting every term by one to the right. It's not clear that shifting terms in series keeps the sum the same. For instance, re-arranging infinitely many terms in conditionally convergent infinite series changes the sum.

> Adding 0 to the front of an infinite series ... not clear that ... keeps the sum the same

I don't know if I would go that far... but I agree with the general spirit of your comment. Which is also the point of my original post. If you think that my appending a zero calls my proof into question, the proof presented in the video takes far more dubious and horrific liberties.

I don't know for certain if (0 + S1) ?= S1, but infinite series require care. Consider this:

Let S1 = 0 + 0 + 0 + ... = 0 Then surely, S1 = (1 - 1) + (1 - 1) ... = 1 - 1 + 1 - 1 + ... -1 + S1 = -1 + 1 - 1 + 1 - 1 ... = (-1 + 1) + (-1 + 1) ... = 0

But then -1 + 0 = 0

Your error is here, equality is wrong: -1 + 1 - 1 + 1 - 1 ... = (-1 + 1) + (-1 + 1) ...

In first series, there are 2 different elements (1 and -1) and series can end at any one of them rendering the end result of the sum uncertain. On second one there is only one element - (-1 + 1) which is 0, so wherever you end it the result is always the same.

My whole point was there are operations that work in finite mathematics that don't work on infinite series, so yes, I didn't prove mathematics inconsistent, I just proved grouping is illegal for divergent infinite series :)

To be clear, yes

(1 - 1) + (1 - 1) + ... != 0 + 0 + ... either

> To be clear, yes

> (1 - 1) + (1 - 1) + ... != 0 + 0 + ... either

I don't see why. I understand the sets themselves are not equal, but the sum of the elements of those sets is at any given index.