# Non-converging Taylor series of $1/(1-x)$

From my own calculations with Maclaurin series and double-checking online, I get the result that:

$${1 \over 1 - x} = \sum_{n=0}^\infty x^n$$

This seems to be true for $-1 \lt x \lt 1$, but for values outside this range, I get some peculiar results. For instance, when try to put $x=2$ in the equation, I get:

$${1 \over 1 - 2} = -1 \\ \sum_{n=0}^\infty 2^n = 1 + 2 + 4 + 8 + 16 + \cdots = \text{indeterminate} \\$$

Why does this equality only apply in this narrow range, even though $1 \over 1 -x$ is defined for any $x \ne 1$?

• In other words, why do analytic functions have series that only converge in a disc/abscissa instead of globally?
– anon
Commented Feb 10, 2012 at 22:30
• Informally, we have to assign meaning to expressions like $\sum_0^\infty x^n$. That meaning is that the limit of the partial sums exists. If $|x|\ge 1$, that limit does not exist. We could define the infinite sum to be $1/(1-x)$ if $x\ne 0$. However, then the normal rules of algebra can break down. Commented Feb 10, 2012 at 22:37

There are Summation Methods for series that, under the usual definition of convergence, do not converge. Some of these methods have applications.

The list of such summation methods is very long: Euler summation, Cesaro summation, Abel summation, Borel summation, many others. I have mentioned these so that you will see that mathematicians of the first rank have considered this kind of problem.

Some methods actually do assign sum $-1$ to $1+2+4+\cdots$. The ordinary rule of algebra that the sum of positive numbers is positive breaks down for any such summation method. Thus, if we use such a summation method, we must be very careful. By way of contrast, doing what comes naturally is ordinarily correct with summation as defined in the usual way.

The field is very large. For a beginning, you may want to start here and chase down some of the references.

You get that formula by taking a limit:

$$\lim_{n\to \infty}\frac{1-x^n}{1-x},$$

where the $x$ comes from the finite sum

$$1+x+x^2+x^3+\cdots +x^n.$$

But, this limit exists only when $-1<x<1$. I believe this is sometimes known as Gauss's sum formula. So, your equality only applies in these cases.

Because the function blows up---i.e. approaches $\infty$---as $x$ approaches $1$. The center about which you're expanding is $0$, i.e. you've got $\sum\limits_{n=0}^\infty (x-0)^n$, and the distance from the center to the nearest point in the complex plane where the function blows up is the radius of convergence. Get farther from the center than that and the series diverges.

I think all previous answers are quite enough, but I guess this will help you understand it too. If we look at $$y = \frac{1}{1+x^2}$$ we can show that it's expansion is

$$y = \sum (-x)^{2k}$$

But then we think.... why is it that this sum is only convergent for $|x|<1$ if the function is everywhere continuous and differentiable? Why does this sum behave this way when the function as such a good behaviour, in contrast to $(1-x)^{-1}$ or $(1+x)^{-1}$?

If we think in a "complex" way, we see that

$$y = \frac{1}{1+z^2}$$ doesn't behave that well. Indeed, it has poles at $z = i$ and $z=-i$. Thus when we think about the unit circle $|z| < 1$ things "make sense".

You can read a little about a result due to Abel:

If $\sum a_n z^n$ converges for $z \neq 0 = z_1$ then

1. It converges absolutely for all $z$ such that $|z| < |z_1|$.
2. It converges unformly in all circular disk with center at $0$ and radius $r<|z_1|$
3. It diverges for all $|z| > |z_2|$ if it diverges for $z_2$.

The theorem reduces to intervals when we deal with real numbers. (consult Apostol's Calculus Chapter 11 for more info.)

Hope this helps.