Maclaurin series of $\frac{1}{(2-x)^2}$

Question

When googling about Maclaurin series, I found this: $$\begin{eqnarray*} \frac{1}{1-x} = \sum_{n=0}^\infty x^n \\ \frac{1}{x} = \frac{1}{1-(1-x)} = \sum_{n=0}^\infty (1-x)^n \\ \frac{1}{x^2} = \sum_{n=0}^\infty (1-x^2)^n \\ \frac{1}{(2-x)^2} & = & \sum_{n=0}^\infty (1-(2-x)^2)^n) \\ & = & \sum_{n=0}^\infty (-x^2 +4x - 3)^n \\ & = & \sum_{n=0}^\infty (-1)^n (x - 3)^n (x - 1)^n \end{eqnarray*}$$

There appears to be problem regarding the convergence at $x = 0$. I don't see where the mistake in the working is, and then how to continue until we get the standard answer for power series of $\frac{1}{(2-x)^2}$? Thank you

Answer 1

That series from the starting is valid if $$|x|<1$$

And you got the hint now.

Find the convergence interaiew..usign this from the starting..and you will see why X=0 gives divergence result

At last you should get $$|1-(2-x)^2|<1$$ Here is the graph showing the above region:

which on solving gives the approximate solution interval as $$0.58<x<2\;\;\;\; \cup \;\;\;\;2<x<3.41$$

So,

$x=0$ not lies in this interval

Answer 2

I would like to suggest another (standard) approach which seems much easier:

$$\frac{1}{(2-x)^2} = \left(\frac{1}{2-x}\right)'$$ $$\frac{1}{2-x} = \frac{1}{2}\cdot \frac{1}{1-\frac{x}{2}}=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^n}{2^n}$$ $$\Rightarrow \frac{1}{(2-x)^2} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{nx^{n-1}}{2^n} = \sum_{n=0}^{\infty}\frac{(n+1)x^{n}}{2^{n+2}}$$

Answer 3

Here is a big hint go get you started.

$\frac {1}{(2-x)^2} = \frac{d}{dx} \frac {1}{2-x}$