Q:

Sequence of $a_n=(-\frac{1}{2})^n$

Considering the sequence of $a_n=(-\frac{1}{2})^n$ where n is a natural number, I have to prove that it is a sequence in which the positive terms and negative terms generate two geometric illiminated progressions having the same ratio then calculate the sum of the terms of $a_n$.
The final part is easy but I don't know how to give a demonstration of what the exercise is asking. I can find some terms then take the positive one and the negative one and finally prove that the two sequences I found are two geometric progression having the same ratio... But I find not very elegant... Can you help me please?
Thanks in advance!
(following my solution the ratio is 1/4) 

A:

Positive terms go with even $\;n\;$:
$$\left\{\frac1{2^{2n}}\right\}_{n\in\Bbb N}=\left\{\frac14\,,\,\frac1{16}\,,\,\frac1{64}\,,\ldots\,,\frac1{4^n}\,,\ldots\right\}$$
The above geometric sequence's constant quotient is
$$\frac{\frac1{4^{n+1}}}{\frac1{4^n}}=\frac14$$
and thus the sum of this series (why does the sum exist?) is
$$\frac{\frac14}{1-\frac1{4}}=\frac13$$
the negative terms with odd $\;n\;$:
$$\left\{-\frac1{2^{2n-1}}\right\}_{n\in\Bbb N}=\left\{-\frac12\,,\,-\frac18\,,\,-\frac1{32}\,,\ldots\,,\,-\frac2{2^{2n}}\,,\ldots\right\}$$
and its quotient is
$$\frac{-\frac2{2^{2(n+1)}}}{-\frac2{2^{2n}}}=\frac14$$
and its sum is
$$\frac{-\frac12}{1-\frac14}=-\frac23$$

