# Could someone explain this contradiction to me?

Standard

Reading on with Julian Havil’s Gamma: Exploring Euler’s Constant and inspired by his discussion of the harmonic series, I come across this:

$\frac{1}{1-e^x} = 1 + e^x + e^{2x} + e^{3x} + ...$

Havil calls this a “non-legitimate binomial expansion” and it seems to me it can be generalised:

$(1 - r^x)^{-1}= 1 + r^x + r^{2x} + r^{3x} + ...$

as $1 = (1 - r^x)(1 + r^x +r^{2x}+r^{3x}+... )= 1 + r^x +r^{2x}+r^{3x}+...-r^x-r^{2x}-r^{3x}-...$

And, indeed if we take $x=-1, r=2$ we get:

$\frac{1}{1-2^{-1}} = 2 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +...$ at the limit.

But if we have $x \geq 0$ it is divergent and the identity, which seems algebraically sound to me, breaks down. E.g., $r=2, x=2$:

$\frac{1}{1-4} = -\frac{1}{3} = 1 + 4 + 8 + 16 + ...$

So what is the flaw in my logic?

## 4 thoughts on “Could someone explain this contradiction to me?”

1. First, the identity works for x > 0 if r has magnitude less than 1. Try r=0.5, x=1.

As to why it breaks down for x >= 0 and |r| >= 1, the right side is mathematical shorthand for the limit of the n-th partial sum as n goes to infinity. If you are raising things equal to or larger than 1 in magnitude to progressively larger powers, the partial sums don’t converge, so the right side is either infinity or didn’t exist, depending on your religious affiliation.

2. Paul Rio says:

Serious paper about the P versus NP Problem

Abstract:$UNIQUE \ SAT$ is the problem of deciding whether a given Boolean formula has exactly one satisfying truth assignment. The $UNIQUE \ SAT$ is $coNP-hard$. We prove the $UNIQUE \ SAT$ is in $NP$, and therefore, $NP = coNP$. Furthermore, we prove if $NP = coNP$, then some problem in $coNPC$ is in $P$, and thus, $P = NP$. In this way, the $P$ versus $NP$ problem is solved with a positive answer.

• Thanks, not much to do with the subject in hand but I’ll leave this here for now.