# Don't understand why this binomial expansion is not valid for x > 1

today I'm studying binomial expansion and I'm a little confused about when certain expressions are valid.

E.g. take this solution from my textbook:

I understand that $(1-x)^{-1}$ has an infinite expansion. I also understand that it is invalid when $x=1$ as $0^{-1}$ is undefined.

However, could someone explain to me why this is invalid for $x>1$? What is invalid about $(1-2)^{-1}$ i.e. $-\frac{1}{1} = -1$?

Thanks!

Edit: Also take the next question. I incorrectly guessed it would be valid for all x greater than or equal to 0, as you can take the square root of any positive number greater than 0. So why is this valid when $\mod{x} < 1$?

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Notice that if $x>1$, then the summation $$1+x+x^2+x^3+\ldots$$ diverges to $\infty$. And thus cannot be equal to a finite number.

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The reason is that if $|x|\ge1$, the series $1+x+x^2+\cdots$ does not converge because its general term does not converge to zero. One needs to be Euler to manipulate such objects... see How Euler did it: divergent series for some explanations. This applies to your second example as well.

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+1 Very interesting link – user7530 May 4 '11 at 18:08
@user: Thanks. To explore (much) further this world of divergent series, a place to start might be mathoverflow.net/questions/19201. – Did May 4 '11 at 18:19

Your observation that the series cannot converge for $x=1$ because $1/(1-x)$ has a singularity at that point is a good one. Your question of why the series cannot converge away from $x=1$ is also good. The answer is that if a power series converges for $x=x_0$ then it converges for $|x|\lt|x_0|$. So, if it converged for say $x=2$ then it would have to converge for $x=1$.

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ah! That makes it somewhat clearer! Thank you. – Danny King May 5 '11 at 10:17

For the question why $(1-2)^{-1}=-1$ but the series $\sum_{n=1}^\infty x^n$ diverge, you may want to look at the page on "analytic continuation". The idea is the following, there are many different power series that converge to $1/(1-x)$. For example:

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ which converge in a disk centered at $x=0$ of radius $1$.

$$\frac{1}{1-x}=\frac{i}{1-i(x-(1+i))}= i\sum_{n=0}^\infty i^n(x-(1+i))^n$$ which converge in a disk centered at $x=1+i$ of radius $1$.

$$\frac{1}{1-x}=-\frac{1}{1+(x-2)}=-\sum_{n=0}^\infty (2-x)^n$$ which converge in a disk centered at $x=2$ of radius $1$.

We note the three disks overlap (draw these three disks on a piece of paper).

Now let's say you start with the 3 power series on the right hand and pretend that we know nothing about the left hand side, so we only know 3 power series, and each of them gives a function on the disk on which they converge, but nothing outside their own disks. Of course, since they give rise to the same functions on the overlapping, we can "patch" them together to get a function on the union of the three disks. If I want to find the value of this function at $x=2$, I just plug it in the third power series, and get the value $-1$. In a sense, by looking at the power series $\sum_{n=0}^\infty x^n$, we start with a function that's only defined inside $\lvert x\rvert<1$, but we can "analytically continue" to extend its domain until it covers $x=2$ by writing down other power series and require they match on the overlapping of the disks of convergence.

Now the natural question arises, could there be more than one way to do this? Indeed, if there are, then that will give us a different functions on the union of disks, and probably a different value at $x=2$. In other words, can we write down another power series (say converge inside the disc $\lvert x-(1+i)\rvert<1$) such that it defines the same function as $\sum_{n=0}^\infty x^n$ on the overlapping of the disks, but differs from the second power series above on the rest of the disc $\lvert x-(1+i)\rvert<1$ away from the overlapping? The theory analytic continuation says no. So in a sense, the power series $\sum_{n=0}^\infty x^n$ not only determines an analytic function in the disc $\lvert x\rvert<1$, but also beyond it. Now you apply the theory again, that tells you that the last power series is uniquely determined by the second, which in turn is uniquely determined by the first. So the value you get for the value at $x=2$ is actually unique by the process. It does not mean that the first power series converge to $-1$ at $x=2$, but it means the function that's determined by the power series,(which secretly, is $1/(1-x)$, but in general we won't have a closed form for other series) has value $-1$ at $x=2$.

Last, there is a problem that our choice of the disks seems to be arbitrary. For example, we can draw many overlapping discs that loops around $x=1$ then comeback to $x=2$. In this instance, they don't cause trouble, but a lot times they do. I'll refer to the Mathworld webpage for example again.

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Euler used the same reasoning as you to derive the "equality" $1 – 2 + 3 – 4 + … = \frac{1}{4}$.

According to Euler,

Let us say, therefore, that the sum of any infinite series is the finite expression, by the expansion of which the series is generated. In this sense the sum of the infinite series $1 − x + x^2 − x^3 + ...$ will be $\frac{1}{1+x}$, because the series arises from the expansion of the fraction, whatever number is put in place of x. If this is agreed, the new definition of the word sum coincides with the ordinary meaning when a series converges; and since divergent series have no sum in the proper sense of the word, no inconvenience can arise from this new terminology. Finally, by means of this definition, we can preserve the utility of divergent series and defend their use from all objections.

I wrote a bit more about this in a blog article a few years ago, though even I admit it's a bit dry.

There is in fact more than one accepted way to assign a value to some divergent series: See for example Cesàro summation, Abel summation, Euler summation, Borel summation, etc.

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Apply ratio test for the binomial expansion terms and u will understand that for convergence(tending to some finite value),the ratio of the consecutive terms should be less than 1.Hence,|x|<1.

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