Binomial expansion of negative exponents.

Question

Let's say I have to expand $(1+x)^{-1}$ using binomial expansion.

Using the theorem, I get:

$$(1+x)^{-1} = 1-x+x^2-x^3+x^4-x^5+x^6+....+{\infty}$$

Substituting $x$ for $1$, I get:

$$\frac{1}{2}= 1-1+1-1+1-1+1+....+{\infty}$$

A similar result arises with higher power of the exponent

For $(1+x)^{-2}$ we get:

$$(1+x)^{-2} = 1-2x+3x^2-4x^3+5x^4-6x^5+7x^6+....+{\infty}$$

Substituting $x$ for $1$, I get:

$$\frac{1}{4}= 1-2+3-4+5-6+7+....+{\infty}$$

How does this makes sense? Help please!

Answer 1

The equality $$\sum_{n=0}^\infty x^n = \frac{1}{x+1}$$

is only true if $|x<1|$, so you cannot substitute $x=1$ into it and expect the result to work.

Answer 2

Some of the most widely misunderstood ideas in mathematics are to do with divergent series and about when it's applicable to assign a value to a series (see all the nonsense on this site about $-1/12$ everything else).

The point is: certain equalities only hold in certain places.

Imagine you've got a number line. Sometimes it's only valid to do a particular operation in one part of the number-line and outside of this region, it doesn't make sense. Your question is an example of this.

And yes, it's true that $1-\frac12+\frac14-\frac18+\ldots=\frac{2}{1+2}=\frac23$ but this doesn't necessarily generalise * to $1-2+4-8+\ldots$. Clearly the first series gets closer to $0.666\ldots$ as you add more terms but the second series doesn't get closer to anything, it just keeps going to $\pm\infty$. So the first series is called convergent (since it approaches something) and the second is called divergent (since it doesn't).

To be explicit, the series $1-x+x^2-x^3+\ldots$ is only equal to anything ** when $x$ is between $-1$ and $1$ (not including the endpoints $1$ or $-1$). In essence, the value of $x$ needs to be small enough so that when you add more terms in the series, the terms get smaller and smaller until they're effectively $0$.

There needs to be a limit of the series for it to be equal to anything. i.e. there needs to be a finite value that the series gets closer to.

Does this make sense?

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For completeness, there are alternative definitions of summation (Abel, Cesaro, analytic continuation) that do allow for summation in places where it's normally undefined. But this is not the normal definition.

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* it doesn't generalise from $x=2$ to $x=-2$

** only equal to any finite number, using normal (not Abel) summation

Answer 3

As the other answers and comments point out, the problem here is "divergent" series. A divergent series is hard to pin down, because it breaks most of the usual rules of arithmetic, like associativity and commutativity of addition.
The series you wrote could be evaluated like this:

$$1 - 1 + 1 -1 +\cdots = (1-1) + (1-1) + (1-1) + \cdots = 0+0+0+\cdots = 0$$

or like this:

$$1 - 1 + 1 -1 +\cdots =1+(-1+1) + (-1+1) + \cdots = 1+0+0+\cdots =1.$$

The associative property says that we can group the additions as we please without changing the result. Yet, when we do it with this divergent series, the result does change. Or we could apply commutativity:

$$1 - 1 + 1 -1 +\cdots = 1 -1 +1+1 -1 +1+1+1 -1 + \cdots = 0+1+2+3+\cdots =\infty.$$

By rearranging and regrouping, we can make your sum add up to any integer we want. Which means it doesn't really have a value. Which in turn means that it doesn't mean anything when you set it equal to $1/2.$