Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$.

Question

$$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$

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 $$\frac{1}{2},\frac{1}{6},\frac{1}{24}, ...?$$

Answer 1

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 $x-1$.

Answer 2

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

Answer 3

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,

$$\begin{align}\frac{x^n-1}{x-1} &= \sum_{j=1}^{n}{n\choose j}(x-1)^{j-1} \\&= {n\choose 1}+{n\choose 2}(x-1)^{1}+{n\choose 3}(x-1)^{2} +{n\choose 4}(x-1)^{3}+\cdots \\&= n +\color{blue}{\frac{n(n-1)}{2}}(x-1)^{1}+\color{blue}{\frac{n(n-1)(n-2)}{6}}(x-1)^{2} +\color{blue}{\frac{n(n-1)(n-2)(n-3)}{24}}(x-1)^{3}+\cdots \end{align}$$