# Sum of x^-n

• Apr 13th 2012, 09:49 AM
jj451
Sum of x^-n
I'm struggling with the following sum:

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

I've tried using a geometric series with the nth term equaling $ar^{n-1}$ where $a,r = \frac{1}{x}$, and using the geometric sum of the first N+2 terms:

$S_{N+2} = \frac{1-\frac{1}{x^{N+2}}}{x-1}$

but I don't think thats correct. I just checked in Wolfram Alpha and the answer there is mine multiplied by $x$ so I'm probably just missing something simple.

Not really sure what to do really so any help would be appreciated.
• Apr 13th 2012, 10:17 AM
Prove It
Re: Sum of x^-n
Quote:

Originally Posted by jj451
I'm struggling with the following sum:

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

I've tried using a geometric series with the nth term equaling $ar^{n-1}$ where $a,r = \frac{1}{x}$, and using the geometric sum of the first N+2 terms:

$S_{N+2} = \frac{1-\frac{1}{x^{N+2}}}{x-1}$

but I don't think thats correct. I just checked in Wolfram Alpha and the answer there is mine multiplied by $x$ so I'm probably just missing something simple.

Not really sure what to do really so any help would be appreciated.

I don't think you have applied the formula correctly...

\displaystyle \begin{align*} \sum_{k = 0}^{N+1}\frac{1}{x^k} = \sum_{r = 0}^{N+1}\left(\frac{1}{x}\right)^k \end{align*}

This is a geometric series with \displaystyle \begin{align*} a = 1, r = \frac{1}{x} \end{align*} and \displaystyle \begin{align*} n = N+2 \end{align*}, so applying the sum formula \displaystyle \begin{align*} S_n = \frac{a\left(1 - r^{n+1}\right)}{1 - r} \end{align*} we have

\displaystyle \begin{align*} S_{N+1} &= \frac{1\left[1 - \left(\frac{1}{x}\right)^{N+2}\right]}{1 - \frac{1}{x}} \\ &= \frac{1 - \left(\frac{1}{x}\right)^{N+2}}{1 - \frac{1}{x}} \\ &= \frac{\frac{x^{N+2} - 1}{x^{N+2}}}{\frac{x - 1}{x}} \\ &= \frac{x\left(x^{N+2} - 1\right)}{x^{N+2}\left(x - 1\right)} \\ &= \frac{x^{N+2} - 1}{x^{N+1}\left(x - 1\right)} \end{align*}
• Apr 13th 2012, 10:23 AM
jj451
Re: Sum of x^-n
Yeah thanks that was kind of dumb of me. I was just staring at it and couldn't make sense of it. Thanks again.