# 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.

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Did you notice that $|x|$ also turned into $x$? The two changes are related.... – Gerry Myerson Nov 6 '12 at 6:21
@GerryMyerson Oh yes I noticed, and understand that for absolute values you have to break it up into the left and right hand side but I still just don't fully understand why the left side flips to 2 and the right stays the same – Richard Nov 6 '12 at 6:24
How is $x$ related to $k$? – Daryl Nov 6 '12 at 6:25
$(\frac 12)^{-k}=2^k$. – Hagen von Eitzen Nov 6 '12 at 6:25
@HagenvonEitzen oh wow that totally makes sense. So you break the integral into two summations, one is $\frac{1}{2}^{-k}$ and the other one is $\frac{1}{2}^k$ but the $-k$ flips, got it. But why the negative out in front of the whole summation? – Richard Nov 6 '12 at 6:28

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 $y=2$.

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uhh...this really didn't make much sense to me....I get the flipping and all, but I just don't get why the negative out front on the $(-1)$ here: $-(-1)+2$ – Richard Nov 6 '12 at 6:55
Well, $$\frac1{1-2}=\frac1{-1}=-1,$$ right? Then $$-\frac1{1-2}=-(-1).$$ – Cameron Buie Nov 6 '12 at 7:01
Does my answer make things easier for you than the initial approach? If it makes you feel any better, I was personally mystified by the jump from $$\sum_{k=-\infty}^{-1}2^k$$ to $$-\frac1{1-2}.$$ I mean, I see that it's true, but it seems like a lot of steps got skipped in there.... I'll go ahead and add in a justification for that in my own answer. – Cameron Buie Nov 6 '12 at 7:05
What do you think, @Richard? Does that make things any more clear for you? – Cameron Buie Nov 6 '12 at 7:45

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 $1$.

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.