Simple change of boundaries of summation

Question

I need to change the boundaries of a summation to get to the following result:

$\sum\limits_{n=-N}^{-1}x^{-n}=\frac{x^{-N}-1}{1-x}$.

Now I know that a geometric series has the following property: $\sum\limits_{n=0}^{\infty}x^{n}=\frac{1-x^{n+1}}{1-x}$.

I just don't seem to get there. I tried the following: $\sum\limits_{n=-N}^{-1}x^{-n}=-1+\sum\limits_{n=0}^{N}x^{-n}$, but how to continue...

Answer 1

We have that

$$\sum\limits_{n=0}^{N}x^{n}=\frac{1-x^{N+1}}{1-x} \implies \sum\limits_{n=1}^{N}x^{n}=\frac{1-x^{N+1}}{1-x}-1=\frac{x-x^{N+1}}{1-x}=x^{N+1}\frac{x^{-N}-1}{1-x}$$

and then

$$\sum\limits_{n=1}^{N}\frac{x^{n}}{x^{N+1}}=\frac{x^{-N}-1}{1-x}$$

$$\sum\limits_{n=1}^{N}x^{n-N-1}=\sum\limits_{n=-N}^{-1}x^{n}=\frac{x^{-N}-1}{1-x}$$

Answer 2

We need to show $$\sum\limits_{n=-N}^{-1}x^{n}=\frac{x^{-N}-1}{1-x}$$

Note that $$\sum\limits_{n=-N}^{-1}x^{n} = x^{-N} + x^{-N+1} +... + x^1$$

$$= x^{-N}(1+x+x^2+...+x^{N-1}) = x^{-N} (\frac {1-x^N}{1-x}) = \frac {x^{-N}-1}{1-x}$$