<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] hint
Let $f (x)=\frac {x^n-1}{x-1}$.
we expand $f (x+1) $ near zero as
$$f (x+1)=((x+1)^n-1)/x=$$
$$(1+nx+\frac {n (n-1)}{2!}x^2+\frac{n (n-1)(n-2)}{3!}x^3+...-1)/x $$
$$=n+\frac {n (n-1)}{2!}x+.... $$
to get the expansion arround $x=1$, replace $x $ by $x-1$.
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