<s>[INST] Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
 [/INST] Hint:
Try this:
$$\frac{1}{2-x} = \frac{\frac{1}{2}}{1-\frac{x}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = \sum_{n=0}^\infty 2^{-(n+1)}x^n$$
Your method gives the Taylor expansion around $x=1$ with radius of convergence of $|x-1| < 1$ so for $x = 0$ you have $|x-1| = |-1| = 1 \not < 1$ you are on the boundary of the radius of convergence and you could end up with a delicate case.
$$\frac{1}{2-x} = \frac{1}{1-(x-1)} = \sum_{n=0}^\infty (x-1)^n = \sum_{n=0}^\infty\sum_{k=0}^n \binom{n}{k}(-1)^{n-k}x^k \\ \stackrel{*}{=} \lim_{m\to\infty} \sum_{n=0}^m \binom{n}{0}(-1)^n + \sum_{n=1}^m \binom{n}{1}(-1)^{n-1}x + \ldots + \sum_{n=k}^m \binom{n}{k}(-1)^{n-k}x^k + \ldots$$
$(*)$ is alowed when we have convergence (wich we don't  at $x=0$ as we will see). Note that at $x=0$  the first sum is divergent and the other ones are $0$. So you can see that we don't have convergence for $x=0$ in the Taylor expansion around $x=1$.
Note that the Taylor expansion at $x=0$ is often called Maclaurin series. If that's what you wanted then the method at the top is the one to go, otherwise, your method is a correct Taylor expansion, but centered at $x=1$.
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