1
$\begingroup$

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...

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

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

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

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

$\endgroup$
  • $\begingroup$ 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 : $-1+\sum_{n = 0}^N x^n = \frac{-x^{N+1}+x}{1-x}$ $\endgroup$ – Thinking Oct 17 '18 at 20:40
1
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.