Task:
Calculate the Taylor expansion including the remainder for the function $f:D\to\mathbb{R},D\subseteq \mathbb{R}$$f(x)=\frac{1}{1-ax}$ at $x^*=0$.
Solution:
I've calculated the first four derivatives first. From these I was able to derive the $n$-th derivative:
$$f^{(n)}(0)=n!\cdot a^n$$
Then I inserted it into the Taylor polynomial with expansion point 0 I discovered that the polynomial looks like this:
$$T_{0,n}(x)=1+ax+a^2x^2+a^3x^3+\dots+a^nx^n+Remainder$$
Now I ask myself the question, how I determine the remainder. Can you help me with that? The remainder-form is: $R_n(x)=f(x)-T_{x^*,n}(x)$
Hint:
Use the high school identity: $$1-u^{n+1}=(1-u)(1+u+u^2+\dots+u^n),$$ whence \begin{align} 1+u&+u^2+\dots+u^n=\frac{1-u^{n+1}}{1-u}=\frac{1}{1-u}-\frac{u^{n+1}}{1-u},\\[0.5ex] \text{so that }\qquad&\frac{1}{1-u}=1+u+u^2+\dots+u^n+\frac{u^{n+1}}{1-u}. \end{align}