Prove $ 1+\left( 1-\frac{x}{n}\right)+...+\left(1-\frac{x}{n}\right)^{n-1}=\frac{n}{x}\left(1-\left(1-\frac{x}{n}\right)^n\right)$

Question

Prove:

$$ 1+\left( 1-\frac{x}{n}\right)+...+\left(1-\frac{x}{n}\right)^{n-1}=\frac{n}{x}\left(1-\left(1-\frac{x}{n}\right)^n\right)$$

The first part could be rewritten as a series such as:

$$\sum_{n=0}^{n-1}\left(1-\frac{x}{n}\right)^n$$

Maybe it could be transform into a Taylor series, but I just don't know where to start.

Answer 1

This is just a sum of terms in geometric progression, mimicking its proof write \begin{align} S&=1+\left( 1-\frac{x}{n}\right)+...+\left(1-\frac{x}{n}\right)^{n-1}\\ S\left( 1-\frac{x}{n}\right)&=\left( 1-\frac{x}{n}\right)+\left(1-\frac{x}{n}\right)^{2}+...+\left( 1-\frac{x}{n}\right)^n \end{align} then subtract second equality from first to get $$ S\frac{x}{n}=1-\left(1-\frac{x}{n}\right)^n $$ and the result follows.

Answer 2

$$\sum_{i=0}^{n-1}\left(1-\frac{x}{n}\right)^i=\frac{1-\left(1-\frac{x}{n}\right)^n}{1-\left(1-\frac{x}{n}\right)}=\frac{n}{x}\left[1-\left(1-\frac{x}{n}\right)^n\right]$$

Answer 3

take $q=1-\frac xn$

you may now $s_n=1+q+q^2+q^3+...+q^{n-1}=\frac{1-q^n}{1-q}$

now put $q=1-\frac xn$
so $$1+\left( 1-\frac{x}{n}\right)+...+\left(1-\frac{x}{n}\right)^{n-1}= \\ 1+q+q^2+...+q^{n-1}\\ \to \frac{1-q^n}{1-q}=\frac{1-(1-\frac{x}{n})^n}{1-(1-\frac{x}{n})}$$