Power series expansion of f(x)=1/(1-x) around x=0 and x=-1

Question

For the power series expansion of the function $f(x)$ I worked out the at $x=0$ the power series expansion is $$1(x-0)^n$$ and at $x=-1$ the power series expansion is $$\left(\frac{1}{2^n}+1\right)\cdot(x+1)^n$$ I was hoping to confirm these expansions.

Answer 1

Hint: remember that for $\lvert x\rvert <1$, $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1+x+x^2+x^3+\dots$$

This gives you the power series expansion at $0$. For the one at $-1$, write $x=-1+\varepsilon$ and $$\frac{1}{1-x} = \frac{1}{2-\varepsilon} = \frac{1}{2}\frac{1}{1-\frac{\varepsilon}{2}}$$ and use the same formula.