QUOTE(Flame Haze @ Nov 17 2013, 03:52 PM)
This is complicated
Let s = 1+2+4+8+....
So 2s = 2+4+8+16....
And s-1 = 2+4+8+16+...
Seems to stand true that 2s = s-1
How come s = -1?

QUOTE(reconnaissance @ Nov 18 2013, 02:24 AM)
However, can it be explained this way?;
2s = s-1
2s-s = (s-1)-s
s = -1
There might be a more sinister motive behind the smiley. Well, straight to the point, it's a
mathematical fallacy, where it presents an invalid proof for 3 reasons as explained in the following:
Reason #1: s has no sum.
s = 1 + 2 + 4 + 8 + … + 2^n is the infinite series whose terms are the successive powers of two. It is logically obvious that the sum of ever-growing positive-definite terms does not converge at a negative value or s = −1. In other words, it diverges to infinity, and so in the usual sense it has no sum.
Reason #2: Division-by-zero fallacy.
Let's start from here: 2s = s − 1. We know s is said to tend to infinity (∞). Using the notion of infinite limits, we can generalize that s − 1 becomes extremely close to s as s approaches indefinitely to infinity (∞).
And so, it follows that
2s = s
Divide by the non-zero s
2 = 1
The fallacious result is the implicit assumption that dividing by 0 is a legitimate operation
2 × 0 = 1 × 0
Dividing by zero gives:
2 × (0/0) = 1 × (0/0)
Reason #3: Invalid argument of power series.
Using Maclaurin series expansion, one can show that 1/(1 − x) = 1 + x + x² + x³ + …
If we plug in x = 2, the series becomes s and it sums the terms on the RHS to the finite value of −1.
Unfortunately, the expansion is only VALID for arguments |x| < 1. In plain English, invalid argument yields invalid result.
This post has been edited by Critical_Fallacy: Nov 18 2013, 03:43 AM