Question

What is the interval of convergence of `\sum_{n=0}^{oo} (\frac{1}{x(1-x)})^n`?

Answer 1

`x in (-oo,(1-sqrt5)/2)U((1+sqrt5)/2,oo)`

Explanation:

We can wee that `sum_{n=0}^oo(1/(x(1-x)))^n` is a geometric series with ratio `r=1/(x(1-x))` .

Now we know that geometric series converge when absolute value of the ratio is smaller than 1 :

`|r|<1 iff-1< r<1`

So we must solve this inequality :

`1/(x(1-x))<1 and 1/(x(1-x))> -1`

Let's begin with the first one :

`1/(x(1-x))<1 iff 1/(x(1-x))-(x(1-x))/(x(1-x))<0 iff`

`(1-x+x^2)/(x(1-x))<0`

We can easily prove that the numerator is always positive and the denominator is negetive in the interval `x in(-oo,0)U(1,oo)`.
So this is the solution for our first inequality.

Let's see the second one :

`1/(x(1-x))+(x(1-x))/(x(1-x))>0 iff(1+x-x^2)/(x(1-x))>0`

This inequality hasas solution the interval:

`x in (-oo,(1-sqrt5)/2)U((1+sqrt5)/2,oo)`

So our series converge where this to intervals are both true.

Thus our interval of convergence is :

`x in (-oo,(1-sqrt5)/2)U((1+sqrt5)/2,oo)`