<s>[INST] Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$. $$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$
I thought that at $x=1$ the expansion is:
$$n+n(n-1)(x-1)+n(n-1)(n-2)(x-1)^2+...$$
but the answer is:
$$n+\frac{1}{2}n(n-1)(x-1)+\frac{1}{6}n(n-1)(n-2)(x-1)^2+...$$
Can you explain the origin of $$\frac{1}{2},\frac{1}{6},\frac{1}{24}, ...?$$
 [/INST] By Binomial formula we have
$$x^n =((x-1)+1)^n  = \sum_{j=0}^{n}{n\choose j}(x-1)^j=1+(x-1) \sum_{j=1}^{n}{n\choose j}(x-1)^{j-1}$$
Hence, 
$$\begin{align}\frac{x^n-1}{x-1} &= \sum_{j=1}^{n}{n\choose j}(x-1)^{j-1} \\&= {n\choose 1}+{n\choose 2}(x-1)^{1}+{n\choose 3}(x-1)^{2} +{n\choose 4}(x-1)^{3}+\cdots
\\&= n +\color{blue}{\frac{n(n-1)}{2}}(x-1)^{1}+\color{blue}{\frac{n(n-1)(n-2)}{6}}(x-1)^{2} +\color{blue}{\frac{n(n-1)(n-2)(n-3)}{24}}(x-1)^{3}+\cdots \end{align}$$
</s>