# Simple change of boundaries of summation

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...

• I obtain $\sum\limits_{n=-N}^{-1}x^{n}=\frac{x^{-N}-1}{1-x}$, can you check that? – gimusi Oct 17 '18 at 20:44
• Did you mean $\sum\limits_{n=-N}^{-1}x^{-n}=-1+\sum\limits_{n=0}^{N}x^{{\color{red} +}n}$ ? – Delta-u Oct 18 '18 at 11:58

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

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

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

• Ok, note that there is a mistake in his question. He is talking about : $\sum_{n = -N}^{-1} x^{-n}$ which is just equal to : $-1+\sum_{n = 0}^N x^n = \frac{-x^{N+1}+x}{1-x}$ – Thinking Oct 17 '18 at 20:40

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

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

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