(March 8, 2023 at 4:47 pm)GrandizerII Wrote: I had a look earlier at another thread recently posted here in this section, in which polymath exposed a paradox when doing math with infinite series a certain way.
I'll reiterate this here:
x = 1 + 2 + 4 + 8 + 16 + ...
2x = 2 + 4 + 8 + 16 + 32 + ...
x = 1 + 2x
x - 2x = 1
-x = 1
x = -1
But x is supposed to be a divergent infinite series, so the sum can't amount to a finite number (such as -1)! What gives then?
I've been reading what other people online had to say about this, and many of them say the issue lies with one of the steps involving an "infinity minus infinity" which is indeterminate (in much the same way as 0/0 is indeterminate).
Namely, this step:
x - 2x = 1
I don't think the issue is with that. The answer is only indeterminate when we're dealing with infinity in an abstract sense. But here, we know what numbers are involved exactly in the infinite series, and so there is no issue with subtracting one well-defined infinite series from another.
The problem, in my opinion, is with the second step. We have 2x = 2 + 4 + 8 + 16 + 32 + ...
But 2x could also be 2x = x + x = 1 + 1 + 2 + 2 + 4 + 4 + 8 + 8 + 16 + 16 + ...
If we then move to the next step, it's not certain anymore that x = 1 + 2x.
The lesson to learn here? Doing algebra with infinite series is tricky as fuck. Don't take it for granted that doing so will lead you to a very intuitive answer.
If you have a convergent series, it is easy enough to show that both operations are legitimate and give the same answer.
So, the sum(a_n +b_n )=sum(a_n) + sum(b_n) *provided the series converge*. If they do not converge, they have no value (unless you choose a different notion of convergence--in which case, you need to prove that the algebra works as expected).
And yes, there *are* other ways of defining infinite sums. The problem is that they may or may not respect various aspects of algebra (for example, preserving inequalities, or giving the same result if you put 0's into the sum).
