<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] Let $f(x)=\frac{x^n-1}{x-1}=\sum_{k=0}^{n-1}x^k$ for $x\ne 1$. $\lim_{x->1}f(x)=n$.
</s>