Hi there. I have some doubts with this exercise. I have to make the summation for:

$\displaystyle \displaystyle\sum_{i=-3}^\infty{(-\displaystyle\frac{1}{3})^i}$

I've solved this way:

$\displaystyle \displaystyle\sum_{i=-3}^{\infty}{(-\displaystyle\frac{1}{3})^i}=\displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}+\displaystyle\sum_{i= 1}^{\infty}{(-\displaystyle\frac{1}{3})^{i-1}(-\displaystyle\frac{1}{3})}$

The doubt I've got is about the "negative" part of the summation, I know how to solve the geometric progression, but I'm not sure about the other part. I've proceeded this way:

$\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=1-2+7-20$

This is what I did: $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=s_{-1}+s_{-2}+s_{-3}+s_{0}$

I'm not sure about this, may I should do it this other way?:

$\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=a_{0}+a_{-1}+a_{-2}+a_{-3}$

I mean, taking the values of the sequence for the series for every "i", and making that summation but the accumulation? cause I think it has not much sense what I did before, I mean there is no accumulation for the negative part, right?

Bye there, and thanks for your help.