## Could someone explain this contradiction to me?

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 responses to “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. Thanks, as always

3. Paul Rio

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.

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