Infinite Geometric Sum So I have this infinite sum:
$$\sum_{k=-\infty}^\infty \left(\frac{1}{2}\right)^{|k|}=3 \tag{1}$$
Which breaks down to:
$$\sum_{k=-\infty}^{-1}2^{k} + \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{k} \tag{2}$$
Which then equates to:
$$-\frac{1}{1-2}+\frac{1}{1-\frac{1}{2}}$$
$$=-(-1)+2$$
$$=3$$
But I just don't understand HOW (1) breaks down to (2). Why does the $\frac{1}{2}$ turn into $2$ from $-\infty$ to $-1$?? Other than that, i understand the rest of the solution.
 A: Let's look at this another way for simplicity.
$\begin{eqnarray*}
\sum_{k=-\infty}^\infty\left(\frac12\right)^{|k|} & = & \left(\frac12\right)^0+\sum_{k=-\infty}^{-1}\left(\frac12\right)^{|k|}+\sum_{k=1}^\infty\left(\frac12\right)^{|k|}\\
& = & 1+\sum_{k=-\infty}^{-1}\left(\frac12\right)^{-k}+\sum_{k=1}^\infty\left(\frac12\right)^k\\
& = & 1+2\sum_{k=1}^\infty\left(\frac12\right)^k.
\end{eqnarray*}$
All we did was split it up a bit and reindex. Can you get the rest of the way from there?

Added: Here's a justification for what (to me) is the biggest leap in the originally-posted approach. A basic geometric series result (arguably, the most basic such result) is that for $|x|<1$, we have $$\sum_{k=0}^\infty x^k=\frac1{1-x},$$ so $$\sum_{k=1}^\infty x^k=-x^0+\sum_{k=0}^\infty x^k=-1+\frac1{1-x}=-\frac{1-x}{1-x}+\frac1{1-x}=\frac{-(1-x)+1}{1-x}=\frac{x}{1-x}.$$ Suppose that $|y|>1$, so if we put $x=\frac1y$--that is, $x=y^{-1}$--then $|x|=\left|\frac1y\right|=\frac1{|y|}<1$, so on the one hand, $$\sum_{k=1}^\infty x^k=\sum_{k=1}^\infty\left(y^{-1}\right)^k=\sum_{k=1}^\infty y^{-k}\:\overset{j\,\mapsto\,k}=\:\sum_{j=1}^\infty y^{-j}\;\overset{k\,\mapsto\,-j}=\;\sum_{k=-\infty}^{-1}y^k,$$ and on the other hand, $$\sum_{k=1}^\infty x^k=\frac{x}{1-x}=\cfrac{\frac1y}{1-\frac1y}=\cfrac1{1-\frac1y}\cdot\frac1y=\cfrac1{\left(1-\frac1y\right)y}=\frac1{y-1}=-\frac1{1-y}.$$ Thus, for $|y|>1$, we have $$\sum_{k=-\infty}^{-1}y^k=-\frac1{1-y}.$$ In particular, our "leap" from $\sum\limits_{k=-\infty}^{-1}2^k$ to $-\frac1{1-2}$ is simply the above result, in the special case $y=2$.
A: Things will be simpler if we examine what's going on. First deal with the sum from $k=0$ to $\infty$. There we are just looking at 
$$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots.\tag{$1$}$$
This is a familiar geometric series, with sum $2$.
Now we examine the part that has $k$ negative. Look first at $k=-1$. Then $|k|=1$. So when $k=-1$, we have $\left(\frac{1}{2}\right)^{|k|}=\frac{1}{2}$.
Now look at $k=-2$. Then $|k|=2$. So when $k=-2$, we have $\left(\frac{1}{2}\right)^{|k|}=\frac{1}{2^2}$.
Now look at $k=-3$. Then $|k|=3$. So when $k=-3$, we have $\left(\frac{1}{2}\right)^{|k|}=\frac{1}{2^3}$. And so on.
So the part that has $k$ negative has sum
$$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots.$$
This too is a familiar geometric series, which we can use the usual formula to sum. However, one might as well observe that the sum is $1$ less than the sum in $(1)$. So the sum is $1$.
Add up.
Remark: There is a widespread dislike of negative numbers, which I share. They are so $\dots$ negative. It is often good strategy to work with positives as much as possible. 
