# power series geometric $1/(1-x)^2$

I am told to find the power series representation and the radius of convergence for $1/(1-x)^2$. The solution of the textbook involves differentiating $1/(1-x)$. But when I first saw this question, I thought that since $1/(1-x)^2 = 1/(1-(2x-x^2))$, it is a geometric power series, and therefore, concluded that the radius of convergence was $2^{1/2}$(solving $r= |2x-x^2|<1$). But this made a problem, since when $x=1$, it is not convergent, but from some theorem that says that a power series can only have three possibilities;

1. the series converges only at x=1,

2. the series converges for all x,

3. the series converges for $|x-a|<R$ and diverges for $|x-a|>R$ and if we apply this to the previous thing, then the interval of convergence should be continuous, but in fact, it is not.

Also, from my procedure, I concluded that the interval of convergence is $1-2^{1/2} < x < 1, 1<x<1+ 2^{1/2}$ but the using the textbook`s method, it is $-1<x<1$

What did I do wrong? Is assuming that the interval of convergence(excluding the endpoint) of a power series and its derivative wrong?

• Around what point is the power series to be made and thus the radius of convergence to be found? Jun 11, 2017 at 8:18

You have to be careful: the theory of power series you're quoting is not about powers of anything, but about powers of $x$ (or $x-x_0$ in the general case). With your approach, you get an expansion $$(1-x)^{-2}=\sum^\infty_{n=0}(2x-x^2)^n,$$ and that's convergent whenever $|2x-x^2|<1,$ indeed. But if you want to transform that into a "normal" power series, you'd have to develop $(2x-x^2)^n$ into a sum of powers of $x$ (binomial theorem), receiving a double series. You'd have to reorder it to join equal powers of $x$, and you can do that only if the (double) series is absolutely convergent. That happens if $2|x|+x^2<1,$ giving you the even stronger restriction $|x|<\sqrt{2}-1.$

Well, let me try to explain what you have essentially done to supplement Professor Vector's and Pedro Tamaroff's answer.

First, using Taylor's expansion and the theory of power series, we have $$\frac{1}{1 - y} = \sum_{k = 0}^{\infty} y^k$$ for all $y \in [-1, 1)$.

Now we can substitute $y = 2x - x^2$ and obtain $$\frac{1}{(1 - x)^2} = \sum_{k = 0}^{\infty} (2x - x^2)^k$$ for all $x \in [1 - \sqrt{2}, 1) \cup (1, 1 + \sqrt{2}]$.

As Professor Vector has observed, the RHS of the above equation is not in the right form. To put it into the right form, we have to impose the restriction $|x| < \sqrt{2} - 1$. This gives us $x \in (1 - \sqrt{2}, \sqrt{2} - 1)$. This only tell us that the radius of convergence $R \ge \sqrt{2} - 1$.

The powerseries $1+x+x^2+x^3+\cdots$ converges for $|x|<1$ and represents $(1-x)^{-1}$. Differentiation does not alter the radius of convergence, so the series obtained by term-wise differentiation $1+2x+3x^2+4x^3+\cdots$ converges for $|x|<1$ and represents $(1-x)^{-2}$, the derivative of $(1-x)^{-1}$.